Given Postorder and Inorder traversals, construct the tree.
Examples:
Input:
in[] = {2, 1, 3}
post[] = {2, 3, 1}Output: Root of below tree
1
/ \
2 3Input:
in[] = {4, 8, 2, 5, 1, 6, 3, 7}
post[] = {8, 4, 5, 2, 6, 7, 3, 1}Output: Root of below tree
1
/ \
2 3
/ \ / \
4 5 6 7
\
8
Approach: To solve the problem follow the below idea:
Note: We have already discussed the construction of trees from Inorder and Preorder traversals:
Follow the below steps:
Let us see the process of constructing a tree from in[] = {4, 8, 2, 5, 1, 6, 3, 7} and post[] = {8, 4, 5, 2, 6, 7, 3, 1}:
- We first find the last node in post[]. The last node is “1”, we know this value is the root as the root always appears at the end of postorder traversal.
- We search “1” in in[] to find the left and right subtrees of the root. Everything on the left of “1” in in[] is in the left subtree and everything on right is in the right subtree.
1
/ \
[4, 8, 2, 5] [6, 3, 7]
- We recur the above process for the following two.
- Recur for in[] = {6, 3, 7} and post[] = {6, 7, 3} Make the created tree as right child of root.
- Recur for in[] = {4, 8, 2, 5} and post[] = {8, 4, 5, 2}. Make the created tree the left child of the root.
Note: One important observation is, that we recursively call for the right subtree before the left subtree as we decrease the index of the postorder index whenever we create a new node
Below is the implementation of the above approach:
C++
/* C++ program to construct tree using inorder and postorder traversals */#include <bits/stdc++.h>using namespace std;/* A binary tree node has data, pointer to left child and a pointer to right child */struct Node { int data; Node *left, *right;};// Utility function to create a new nodeNode* newNode(int data);/* Prototypes for utility functions */int search(int arr[], int strt, int end, int value);/* Recursive function to construct binary of size n from Inorder traversal in[] and Postorder traversal post[]. Initial values of inStrt and inEnd should be 0 and n -1. The function doesn't do any error checking for cases where inorder and postorder do not form a tree */Node* buildUtil(int in[], int post[], int inStrt, int inEnd, int* pIndex){ // Base case if (inStrt > inEnd) return NULL; /* Pick current node from Postorder traversal using postIndex and decrement postIndex */ Node* node = newNode(post[*pIndex]); (*pIndex)--; /* If this node has no children then return */ if (inStrt == inEnd) return node; /* Else find the index of this node in Inorder traversal */ int iIndex = search(in, inStrt, inEnd, node->data); /* Using index in Inorder traversal, construct left and right subtrees */ node->right = buildUtil(in, post, iIndex + 1, inEnd, pIndex); node->left = buildUtil(in, post, inStrt, iIndex - 1, pIndex); return node;}// This function mainly initializes index of root// and calls buildUtil()Node* buildTree(int in[], int post[], int n){ int pIndex = n - 1; return buildUtil(in, post, 0, n - 1, &pIndex);}/* Function to find index of value in arr[start...end] The function assumes that value is postsent in in[] */int search(int arr[], int strt, int end, int value){ int i; for (i = strt; i <= end; i++) { if (arr[i] == value) break; } return i;}/* Helper function that allocates a new node */Node* newNode(int data){ Node* node = (Node*)malloc(sizeof(Node)); node->data = data; node->left = node->right = NULL; return (node);}/* This function is here just to test */void preOrder(Node* node){ if (node == NULL) return; printf("%d ", node->data); preOrder(node->left); preOrder(node->right);}// Driver codeint main(){ int in[] = { 4, 8, 2, 5, 1, 6, 3, 7 }; int post[] = { 8, 4, 5, 2, 6, 7, 3, 1 }; int n = sizeof(in) / sizeof(in[0]); Node* root = buildTree(in, post, n); cout << "Preorder of the constructed tree : \n"; preOrder(root); return 0;} |
Java
// Java program to construct a tree using inorder// and postorder traversals/* A binary tree node has data, pointer to left child and a pointer to right child */class Node { int data; Node left, right; public Node(int data) { this.data = data; left = right = null; }}class BinaryTree { /* Recursive function to construct binary of size n from Inorder traversal in[] and Postorder traversal post[]. Initial values of inStrt and inEnd should be 0 and n -1. The function doesn't do any error checking for cases where inorder and postorder do not form a tree */ Node buildUtil(int in[], int post[], int inStrt, int inEnd, int postStrt, int postEnd) { // Base case if (inStrt > inEnd) return null; /* Pick current node from Postorder traversal using postIndex and decrement postIndex */ Node node = new Node(post[postEnd]); /* If this node has no children then return */ if (inStrt == inEnd) return node; int iIndex = search(in, inStrt, inEnd, node.data); /* Using index in Inorder traversal, construct left and right subtrees */ node.left = buildUtil( in, post, inStrt, iIndex - 1, postStrt, postStrt - inStrt + iIndex - 1); node.right = buildUtil(in, post, iIndex + 1, inEnd, postEnd - inEnd + iIndex, postEnd - 1); return node; } /* Function to find index of value in arr[start...end] The function assumes that value is postsent in in[] */ int search(int arr[], int strt, int end, int value) { int i; for (i = strt; i <= end; i++) { if (arr[i] == value) break; } return i; } /* This function is here just to test */ void preOrder(Node node) { if (node == null) return; System.out.print(node.data + " "); preOrder(node.left); preOrder(node.right); } // Driver Code public static void main(String[] args) { BinaryTree tree = new BinaryTree(); int in[] = new int[] { 4, 8, 2, 5, 1, 6, 3, 7 }; int post[] = new int[] { 8, 4, 5, 2, 6, 7, 3, 1 }; int n = in.length; Node root = tree.buildUtil(in, post, 0, n - 1, 0, n - 1); System.out.println( "Preorder of the constructed tree : "); tree.preOrder(root); }}// This code has been contributed by Mayank// Jaiswal(mayank_24) |
Python3
# Python3 program to construct tree using# inorder and postorder traversals# Helper function that allocates# a new nodeclass newNode: def __init__(self, data): self.data = data self.left = self.right = None# Recursive function to construct binary# of size n from Inorder traversal in[]# and Postorder traversal post[]. Initial# values of inStrt and inEnd should be# 0 and n -1. The function doesn't do any# error checking for cases where inorder# and postorder do not form a treedef buildUtil(In, post, inStrt, inEnd, pIndex): # Base case if (inStrt > inEnd): return None # Pick current node from Postorder traversal # using postIndex and decrement postIndex node = newNode(post[pIndex[0]]) pIndex[0] -= 1 # If this node has no children # then return if (inStrt == inEnd): return node # Else find the index of this node # in Inorder traversal iIndex = search(In, inStrt, inEnd, node.data) # Using index in Inorder traversal, # construct left and right subtress node.right = buildUtil(In, post, iIndex + 1, inEnd, pIndex) node.left = buildUtil(In, post, inStrt, iIndex - 1, pIndex) return node# This function mainly initializes index# of root and calls buildUtil()def buildTree(In, post, n): pIndex = [n - 1] return buildUtil(In, post, 0, n - 1, pIndex)# Function to find index of value in# arr[start...end]. The function assumes# that value is postsent in in[]def search(arr, strt, end, value): i = 0 for i in range(strt, end + 1): if (arr[i] == value): break return i# This function is here just to testdef preOrder(node): if node == None: return print(node.data, end=" ") preOrder(node.left) preOrder(node.right)# Driver codeif __name__ == '__main__': In = [4, 8, 2, 5, 1, 6, 3, 7] post = [8, 4, 5, 2, 6, 7, 3, 1] n = len(In) root = buildTree(In, post, n) print("Preorder of the constructed tree :") preOrder(root)# This code is contributed by PranchalK |
C#
// C# program to construct a tree using// inorder and postorder traversalsusing System;/* A binary tree node has data, pointerto left child and a pointer to right child */public class Node { public int data; public Node left, right; public Node(int data) { this.data = data; left = right = null; }}// Class Index created to implement// pass by reference of Indexpublic class Index { public int index;}class GFG { /* Recursive function to construct binary of size n from Inorder traversal in[] and Postorder traversal post[]. Initial values of inStrt and inEnd should be 0 and n -1. The function doesn't do any error checking for cases where inorder and postorder do not form a tree */ public virtual Node buildUtil(int[] @in, int[] post, int inStrt, int inEnd, Index pIndex) { // Base case if (inStrt > inEnd) { return null; } /* Pick current node from Postorder traversal using postIndex and decrement postIndex */ Node node = new Node(post[pIndex.index]); (pIndex.index)--; /* If this node has no children then return */ if (inStrt == inEnd) { return node; } /* Else find the index of this node in Inorder traversal */ int iIndex = search(@in, inStrt, inEnd, node.data); /* Using index in Inorder traversal, construct left and right subtrees */ node.right = buildUtil(@in, post, iIndex + 1, inEnd, pIndex); node.left = buildUtil(@in, post, inStrt, iIndex - 1, pIndex); return node; } // This function mainly initializes // index of root and calls buildUtil() public virtual Node buildTree(int[] @in, int[] post, int n) { Index pIndex = new Index(); pIndex.index = n - 1; return buildUtil(@in, post, 0, n - 1, pIndex); } /* Function to find index of value in arr[start...end]. The function assumes that value is postsent in in[] */ public virtual int search(int[] arr, int strt, int end, int value) { int i; for (i = strt; i <= end; i++) { if (arr[i] == value) { break; } } return i; } /* This function is here just to test */ public virtual void preOrder(Node node) { if (node == null) { return; } Console.Write(node.data + " "); preOrder(node.left); preOrder(node.right); } // Driver Code public static void Main(string[] args) { GFG tree = new GFG(); int[] @in = new int[] { 4, 8, 2, 5, 1, 6, 3, 7 }; int[] post = new int[] { 8, 4, 5, 2, 6, 7, 3, 1 }; int n = @in.Length; Node root = tree.buildTree(@in, post, n); Console.WriteLine( "Preorder of the constructed tree : "); tree.preOrder(root); }}// This code is contributed by Shrikant13 |
Javascript
<script>// Javascript program to construct a tree using inorder// and postorder traversals /* A binary tree node has data, pointer to left child and a pointer to right child */ class Node { constructor(data) { this.data = data; this.left = this.right = null; } } /* Recursive function to construct binary of size n from Inorder traversal in[] and Postorder traversal post[]. Initial values of inStrt and inEnd should be 0 and n -1. The function doesn't do any error checking for cases where inorder and postorder do not form a tree */ function buildUtil(In, post, inStrt, inEnd, postStrt, postEnd) { // Base case if (inStrt > inEnd) return null; /* Pick current node from Postorder traversal using postIndex and decrement postIndex */ let node = new Node(post[postEnd]); /* If this node has no children then return */ if (inStrt == inEnd) return node; let iIndex = search(In, inStrt, inEnd, node.data); /* Using index in Inorder traversal, construct left and right subtrees */ node.left = buildUtil( In, post, inStrt, iIndex - 1, postStrt, postStrt - inStrt + iIndex - 1); node.right = buildUtil(In, post, iIndex + 1, inEnd, postEnd - inEnd + iIndex, postEnd - 1); return node; } /* Function to find index of value in arr[start...end] The function assumes that value is postsent in in[] */ function search(arr,strt,end,value) { let i; for (i = strt; i <= end; i++) { if (arr[i] == value) break; } return i; } /* This function is here just to test */ function preOrder(node) { if (node == null) return; document.write(node.data + " "); preOrder(node.left); preOrder(node.right); } // Driver Code let In=[4, 8, 2, 5, 1, 6, 3, 7]; let post=[8, 4, 5, 2, 6, 7, 3, 1]; let n = In.length; let root = buildUtil(In, post, 0, n - 1, 0, n - 1); document.write( "Preorder of the constructed tree : <br>"); preOrder(root); // This code is contributed by unknown2108</script> |
Output
Preorder of the constructed tree : 1 2 4 8 5 3 6 7
Time Complexity: O(N2), Where N is the length of the given inorder array
Auxiliary Space: O(N), for recursive call stack
Construct a Binary Tree from Postorder and Inorder using hashing:
To solve the problem follow the below idea:
We can optimize the above solution using hashing. We store indexes of inorder traversal in a hash table. So that search can be done O(1) time If given that element in the tree is not repeated.
Follow the below steps to solve the problem:
- We first find the last node in post[]. The last node is “1”, we know this value is the root as the root always appears at the end of postorder traversal.
- we get the index of postorder[i], in inorder using the map to find the left and right subtrees of the root. Everything on the left of “1” in in[] is in the left subtree and everything on right is in the right subtree.
- We recur the above process for the following two.
- Recur for in[] = {6, 3, 7} and post[] = {6, 7, 3} Make the created tree as right child of root.
- Recur for in[] = {4, 8, 2, 5} and post[] = {8, 4, 5, 2}. Make the created tree the left child of the root.
Below is the implementation of the above approach:
C++
/* C++ program to construct tree using inorder andpostorder traversals */#include <bits/stdc++.h>using namespace std;/* A binary tree node has data, pointer to leftchild and a pointer to right child */struct Node { int data; Node *left, *right;};// Utility function to create a new nodeNode* newNode(int data);/* Recursive function to construct binary of size nfrom Inorder traversal in[] and Postorder traversalpost[]. Initial values of inStrt and inEnd shouldbe 0 and n -1. The function doesn't do any errorchecking for cases where inorder and postorderdo not form a tree */Node* buildUtil(int in[], int post[], int inStrt, int inEnd, int* pIndex, unordered_map<int, int>& mp){ // Base case if (inStrt > inEnd) return NULL; /* Pick current node from Postorder traversal using postIndex and decrement postIndex */ int curr = post[*pIndex]; Node* node = newNode(curr); (*pIndex)--; /* If this node has no children then return */ if (inStrt == inEnd) return node; /* Else find the index of this node in Inorder traversal */ int iIndex = mp[curr]; /* Using index in Inorder traversal, construct left and right subtrees */ node->right = buildUtil(in, post, iIndex + 1, inEnd, pIndex, mp); node->left = buildUtil(in, post, inStrt, iIndex - 1, pIndex, mp); return node;}// This function mainly creates an unordered_map, then// calls buildTreeUtil()struct Node* buildTree(int in[], int post[], int len){ // Store indexes of all items so that we // we can quickly find later unordered_map<int, int> mp; for (int i = 0; i < len; i++) mp[in[i]] = i; int index = len - 1; // Index in postorder return buildUtil(in, post, 0, len - 1, &index, mp);}/* Helper function that allocates a new node */Node* newNode(int data){ Node* node = (Node*)malloc(sizeof(Node)); node->data = data; node->left = node->right = NULL; return (node);}/* This function is here just to test */void preOrder(Node* node){ if (node == NULL) return; printf("%d ", node->data); preOrder(node->left); preOrder(node->right);}// Driver codeint main(){ int in[] = { 4, 8, 2, 5, 1, 6, 3, 7 }; int post[] = { 8, 4, 5, 2, 6, 7, 3, 1 }; int n = sizeof(in) / sizeof(in[0]); Node* root = buildTree(in, post, n); cout << "Preorder of the constructed tree : \n"; preOrder(root); return 0;} |
Java
/* Java program to construct tree using inorder andpostorder traversals */import java.util.*;class GFG { /* A binary tree node has data, pointer to left child and a pointer to right child */ static class Node { int data; Node left, right; }; // Utility function to create a new node /* Helper function that allocates a new node */ static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return (node); } /* Recursive function to construct binary of size n from Inorder traversal in[] and Postorder traversal post[]. Initial values of inStrt and inEnd should be 0 and n -1. The function doesn't do any error checking for cases where inorder and postorder do not form a tree */ static Node buildUtil(int in[], int post[], int inStrt, int inEnd) { // Base case if (inStrt > inEnd) return null; /* Pick current node from Postorder traversal using postIndex and decrement postIndex */ int curr = post[index]; Node node = newNode(curr); (index)--; /* If this node has no children then return */ if (inStrt == inEnd) return node; /* Else find the index of this node in Inorder traversal */ int iIndex = mp.get(curr); /* Using index in Inorder traversal, con left and right subtrees */ node.right = buildUtil(in, post, iIndex + 1, inEnd); node.left = buildUtil(in, post, inStrt, iIndex - 1); return node; } static HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); static int index; // This function mainly creates an unordered_map, then // calls buildTreeUtil() static Node buildTree(int in[], int post[], int len) { // Store indexes of all items so that we // we can quickly find later for (int i = 0; i < len; i++) mp.put(in[i], i); index = len - 1; // Index in postorder return buildUtil(in, post, 0, len - 1); } /* This function is here just to test */ static void preOrder(Node node) { if (node == null) return; System.out.printf("%d ", node.data); preOrder(node.left); preOrder(node.right); } // Driver code public static void main(String[] args) { int in[] = { 4, 8, 2, 5, 1, 6, 3, 7 }; int post[] = { 8, 4, 5, 2, 6, 7, 3, 1 }; int n = in.length; Node root = buildTree(in, post, n); System.out.print( "Preorder of the constructed tree : \n"); preOrder(root); }}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to construct tree using inorder# and postorder traversals# A binary tree node has data, pointer to left# child and a pointer to right childclass Node: def __init__(self, x): self.data = x self.left = None self.right = None# Recursive function to construct binary of size n# from Inorder traversal in[] and Postorder traversal# post[]. Initial values of inStrt and inEnd should# be 0 and n -1. The function doesn't do any error# checking for cases where inorder and postorder# do not form a treedef buildUtil(inn, post, innStrt, innEnd): global mp, index # Base case if (innStrt > innEnd): return None # Pick current node from Postorder traversal # using postIndex and decrement postIndex curr = post[index] node = Node(curr) index -= 1 # If this node has no children then return if (innStrt == innEnd): return node # Else find the index of this node inn # Inorder traversal iIndex = mp[curr] # Using index in Inorder traversal, # construct left and right subtrees node.right = buildUtil(inn, post, iIndex + 1, innEnd) node.left = buildUtil(inn, post, innStrt, iIndex - 1) return node# This function mainly creates an unordered_map,# then calls buildTreeUtil()def buildTree(inn, post, lenn): global index # Store indexes of all items so that we # we can quickly find later for i in range(lenn): mp[inn[i]] = i # Index in postorder index = lenn - 1 return buildUtil(inn, post, 0, lenn - 1)# This function is here just to testdef preOrder(node): if (node == None): return print(node.data, end=" ") preOrder(node.left) preOrder(node.right)# Driver Codeif __name__ == '__main__': inn = [4, 8, 2, 5, 1, 6, 3, 7] post = [8, 4, 5, 2, 6, 7, 3, 1] n = len(inn) mp, index = {}, 0 root = buildTree(inn, post, n) print("Preorder of the constructed tree :") preOrder(root)# This code is contributed by mohit kumar 29 |
C#
/* C# program to construct tree using inorder andpostorder traversals */using System;using System.Collections.Generic;class GFG { /* A binary tree node has data, pointer to left child and a pointer to right child */ public class Node { public int data; public Node left, right; }; // Utility function to create a new node /* Helper function that allocates a new node */ static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return (node); } /* Recursive function to construct binary of size n from Inorder traversal in[] and Postorder traversal post[]. Initial values of inStrt and inEnd should be 0 and n -1. The function doesn't do any error checking for cases where inorder and postorder do not form a tree */ static Node buildUtil(int[] init, int[] post, int inStrt, int inEnd) { // Base case if (inStrt > inEnd) return null; /* Pick current node from Postorder traversal using postIndex and decrement postIndex */ int curr = post[index]; Node node = newNode(curr); (index)--; /* If this node has no children then return */ if (inStrt == inEnd) return node; /* Else find the index of this node in Inorder traversal */ int iIndex = mp[curr]; /* Using index in Inorder traversal, con left and right subtrees */ node.right = buildUtil(init, post, iIndex + 1, inEnd); node.left = buildUtil(init, post, inStrt, iIndex - 1); return node; } static Dictionary<int, int> mp = new Dictionary<int, int>(); static int index; // This function mainly creates an unordered_map, then // calls buildTreeUtil() static Node buildTree(int[] init, int[] post, int len) { // Store indexes of all items so that we // we can quickly find later for (int i = 0; i < len; i++) mp.Add(init[i], i); index = len - 1; // Index in postorder return buildUtil(init, post, 0, len - 1); } /* This function is here just to test */ static void preOrder(Node node) { if (node == null) return; Console.Write(node.data + " "); preOrder(node.left); preOrder(node.right); } // Driver code public static void Main(String[] args) { int[] init = { 4, 8, 2, 5, 1, 6, 3, 7 }; int[] post = { 8, 4, 5, 2, 6, 7, 3, 1 }; int n = init.Length; Node root = buildTree(init, post, n); Console.Write( "Preorder of the constructed tree : \n"); preOrder(root); }}// This code is contributed by Rajput-Ji |
Javascript
<script> /* JavaScript program to construct tree using inorder and postorder traversals */ /* A binary tree node has data, pointer to left child and a pointer to right child */ class Node { constructor() { this.data = 0; this.left = null; this.right = null; } } // Utility function to create a new node /* Helper function that allocates a new node */ function newNode(data) { var node = new Node(); node.data = data; node.left = node.right = null; return node; } /* Recursive function to construct binary of size n from Inorder traversal in[] and Postorder traversal post[]. Initial values of inStrt and inEnd should be 0 and n -1. The function doesn't do any error checking for cases where inorder and postorder do not form a tree */ function buildUtil(init, post, inStrt, inEnd) { // Base case if (inStrt > inEnd) { return null; } /* Pick current node from Postorder traversal using postIndex and decrement postIndex */ var curr = post[index]; var node = newNode(curr); index--; /* If this node has no children then return */ if (inStrt == inEnd) { return node; } /* Else find the index of this node in Inorder traversal */ var iIndex = mp[curr]; /* Using index in Inorder traversal, con left and right subtrees */ node.right = buildUtil(init, post, iIndex + 1, inEnd); node.left = buildUtil(init, post, inStrt, iIndex - 1); return node; } var mp = {}; var index; // This function mainly creates an unordered_map, then // calls buildTreeUtil() function buildTree(init, post, len) { // Store indexes of all items so that we // we can quickly find later for (var i = 0; i < len; i++) { mp[init[i]] = i; } index = len - 1; // Index in postorder return buildUtil(init, post, 0, len - 1); } /* This function is here just to test */ function preOrder(node) { if (node == null) { return; } document.write(node.data + " "); preOrder(node.left); preOrder(node.right); } // Driver code var init = [4, 8, 2, 5, 1, 6, 3, 7]; var post = [8, 4, 5, 2, 6, 7, 3, 1]; var n = init.length; var root = buildTree(init, post, n); document.write("Preorder of the constructed tree : <br>"); preOrder(root); </script> |
Output
Preorder of the constructed tree : 1 2 4 8 5 3 6 7
Time Complexity: O(N)
Auxiliary Space: O(N), The extra space is used due to the recursion call stack and to store the elements in the map.
Construct a Binary Tree from Postorder and Inorder using stack and set:
We can use the stack and set without using recursion.
Follow the below steps to solve the problem:
- Create a stack and a set of type Node* and initialize an integer postIndex with N-1
- Run a for loop with p and i, from n-1 to 0
- Create a new Node with value as postorder[p] and set it as the root node, if it is the first node of our newly created tree
- Check if the value of stack top is already present in the set, then remove it from the set and set the left child of stack top equal to the new node and pop out the stack top
- Push the current node into the stack
- Perform step numbers 3,4 and 5 while p is greater than or equal to zero and postorder[p] is not equal to inorder[i]
- Set the new node equal to null and while the stack’s top data is equal to the inorder[i], set the node equal to stack top and pop out the stack top
- If the node is not null then insert the node into the set and push it into the stack also
- Return root of the newly created tree
Below is the implementation of the above idea:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;/* A binary tree node has data, pointerto left child and a pointer to rightchild */struct Node { int data; Node *left, *right; Node(int x) { data = x; left = right = NULL; }};/*Tree building function*/Node* buildTree(int in[], int post[], int n){ // Create Stack of type Node* stack<Node*> st; // Create Set of type Node* set<Node*> s; // Initialise postIndex with n - 1 int postIndex = n - 1; // Initialise root with NULL Node* root = NULL; for (int p = n - 1, i = n - 1; p >= 0;) { // Initialise node with NULL Node* node = NULL; // Run do-while loop do { // Initialise node with // new Node(post[p]); node = new Node(post[p]); // Check is root is // equal to NULL if (root == NULL) { root = node; } // If size of set // is greater than 0 if (st.size() > 0) { // If st.top() is present in the // set s if (s.find(st.top()) != s.end()) { s.erase(st.top()); st.top()->left = node; st.pop(); } else { st.top()->right = node; } } st.push(node); } while (post[p--] != in[i] && p >= 0); node = NULL; // If the stack is not empty and // st.top()->data is equal to in[i] while (st.size() > 0 && i >= 0 && st.top()->data == in[i]) { node = st.top(); // Pop elements from stack st.pop(); i--; } // if node not equal to NULL if (node != NULL) { s.insert(node); st.push(node); } } // Return root return root;}/* for print preOrder Traversal */void preOrder(Node* node){ if (node == NULL) return; printf("%d ", node->data); preOrder(node->left); preOrder(node->right);}// Driver Codeint main(){ int in[] = { 4, 8, 2, 5, 1, 6, 3, 7 }; int post[] = { 8, 4, 5, 2, 6, 7, 3, 1 }; int n = sizeof(in) / sizeof(in[0]); // Function Call Node* root = buildTree(in, post, n); cout << "Preorder of the constructed tree : \n"; // Function Call for preOrder preOrder(root); return 0;} |
Java
// Java program for above approachimport java.io.*;import java.util.*;class GFG { // Node class static class Node { int data; Node left, right; // Constructor Node(int x) { data = x; left = right = null; } } // Tree building function static Node buildTree(int in[], int post[], int n) { // Create Stack of type Node* Stack<Node> st = new Stack<>(); // Create HashSet of type Node* HashSet<Node> s = new HashSet<>(); // Initialise postIndex with n - 1 int postIndex = n - 1; // Initialise root with null Node root = null; for (int p = n - 1, i = n - 1; p >= 0😉 { // Initialise node with NULL Node node = null; // Run do-while loop do { // Initialise node with // new Node(post[p]); node = new Node(post[p]); // Check is root is // equal to NULL if (root == null) { root = node; } // If size of set // is greater than 0 if (st.size() > 0) { // If st.peek() is present in the // set s if (s.contains(st.peek())) { s.remove(st.peek()); st.peek().left = node; st.pop(); } else { st.peek().right = node; } } st.push(node); } while (post[p--] != in[i] && p >= 0); node = null; // If the stack is not empty and // st.top().data is equal to in[i] while (st.size() > 0 && i >= 0 && st.peek().data == in[i]) { node = st.peek(); // Pop elements from stack st.pop(); i--; } // If node not equal to NULL if (node != null) { s.add(node); st.push(node); } } // Return root return root; } // For print preOrder Traversal static void preOrder(Node node) { if (node == null) return; System.out.printf("%d ", node.data); preOrder(node.left); preOrder(node.right); } // Driver Code public static void main(String[] args) { int in[] = { 4, 8, 2, 5, 1, 6, 3, 7 }; int post[] = { 8, 4, 5, 2, 6, 7, 3, 1 }; int n = in.length; // Function Call Node root = buildTree(in, post, n); System.out.print( "Preorder of the constructed tree : \n"); // Function Call for preOrder preOrder(root); }}// This code is contributed by sujitmeshram |
Python3
# Python program for above approach# A binary tree node has data, pointer# to left child and a pointer to right# childclass Node: def __init__(self, x): self.data = x self.left = None self.right = None# Tree building functiondef buildTree(inorder, post, n): # Create Stack of type Node st = [] # Create Set of type Node set = [] # Initialise postIndex with n - 1 postIndex = n - 1 # Initialise root with NULL root = None p = n-1 i = n-1 while p >= 0: # Initialise node with NULL node = None # Run loop while True: # initialize new node node = Node(post[p]) # check if root is equal to null if root == None: root = node # If size of set is greater than 0 if len(st) > 0: # If st top is present in the set s if st[-1] in set: set.remove(st[-1]) st[-1].left = node st.pop() else: st[-1].right = node st.append(node) p -= 1 if post[p+1] == inorder[i] or p < 0: break node = None # If the stack is not empty and st top data is equal to in[i] while len(st) > 0 and i >= 0 and st[-1].data == inorder[i]: node = st[-1] # Pop elements from stack st.pop() i -= 1 # if node not equal to None if node != None: set.append(node) st.append(node) # Return root return root# for print preOrder Traversaldef preOrder(node): if node == None: return print(node.data, end=" ") preOrder(node.left) preOrder(node.right)# Driver Codeif __name__ == '__main__': inorder = [4, 8, 2, 5, 1, 6, 3, 7] post = [8, 4, 5, 2, 6, 7, 3, 1] n = len(inorder) # Function Call root = buildTree(inorder, post, n) print("Preorder of the constructed tree :") # Function Call for preOrder preOrder(root)# This code is contributed by Tapesh(tapeshdua420) |
C#
// C# program for above approachusing System;using System.Collections.Generic;class GFG { // Node class public class Node { public int data; public Node left, right; // Constructor public Node(int x) { data = x; left = right = null; } } // Tree building function static Node buildTree(int[] init, int[] post, int n) { // Create Stack of type Node* Stack<Node> st = new Stack<Node>(); // Create HashSet of type Node* HashSet<Node> s = new HashSet<Node>(); // Initialise root with null Node root = null; for (int p = n - 1, i = n - 1; p >= 0;) { // Initialise node with NULL Node node = null; // Run do-while loop do { // Initialise node with // new Node(post[p]); node = new Node(post[p]); // Check is root is // equal to NULL if (root == null) { root = node; } // If size of set // is greater than 0 if (st.Count > 0) { // If st.Peek() is present in the // set s if (s.Contains(st.Peek())) { s.Remove(st.Peek()); st.Peek().left = node; st.Pop(); } else { st.Peek().right = node; } } st.Push(node); } while (post[p--] != init[i] && p >= 0); node = null; // If the stack is not empty and // st.top().data is equal to in[i] while (st.Count > 0 && i >= 0 && st.Peek().data == init[i]) { node = st.Peek(); // Pop elements from stack st.Pop(); i--; } // If node not equal to NULL if (node != null) { s.Add(node); st.Push(node); } } // Return root return root; } // For print preOrder Traversal static void preOrder(Node node) { if (node == null) return; Console.Write(node.data + " "); preOrder(node.left); preOrder(node.right); } // Driver Code public static void Main(String[] args) { int[] init = { 4, 8, 2, 5, 1, 6, 3, 7 }; int[] post = { 8, 4, 5, 2, 6, 7, 3, 1 }; int n = init.Length; // Function Call Node root = buildTree(init, post, n); Console.Write( "Preorder of the constructed tree : \n"); // Function Call for preOrder preOrder(root); }}// This code is contributed by aashish1995 |
Javascript
// JavaScript program for above approach// Node classclass Node { constructor(data) { this.data = data; this.left = null; this.right = null; }}// Tree building functionfunction buildTree(inorder, postorder, n) { // Create Stack of type Node let st = []; // Create Set of type Node let s = new Set(); // Initialize postIndex with n - 1 let postIndex = n - 1; // Initialize root with null let root = null; for (let p = n - 1, i = n - 1; p >= 0; ) { // Initialize node with null let node = null; do { // Initialize node with new Node(post[p]); node = new Node(postorder[p]); // Check if root is equal to null if (root === null) { root = node; } // If size of set is greater than 0 if (st.length > 0) { // If st[st.length - 1] is present in the set s if (s.has(st[st.length - 1])) { s.delete(st[st.length - 1]); st[st.length - 1].left = node; st.pop(); } else { st[st.length - 1].right = node; } } st.push(node); } while (postorder[p--] !== inorder[i] && p >= 0); node = null; // If the stack is not empty and // st[st.length - 1].data is equal to in[i] while ( st.length > 0 && i >= 0 && st[st.length - 1].data === inorder[i] ) { node = st[st.length - 1]; // Pop elements from stack st.pop(); i--; } // If node not equal to null if (node !== null) { s.add(node); st.push(node); } } // Return root return root;}// For print preOrder Traversalfunction preOrder(node) { if (node === null) return; console.log(node.data + " "); preOrder(node.left); preOrder(node.right);}// Testconst inorder = [4, 8, 2, 5, 1, 6, 3, 7];const postorder = [8, 4, 5, 2, 6, 7, 3, 1];const n = inorder.length;// Function Callconst root = buildTree(inorder, postorder, n);console.log("Preorder of the constructed tree: ");// Function Call for preOrderpreOrder(root);// This code is contributed by lokesh. |
Output
Preorder of the constructed tree : 1 2 4 8 5 3 6 7
Time Complexity: O(N)
Auxiliary Space: O(N), The extra space is used to store the elements in the stack and set.
This article is contributed by Rishi. Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above
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