Given a Binary tree, Traverse it using DFS using recursion.
Unlike linear data structures (Array, Linked List, Queues, Stacks, etc) which have only one logical way to traverse them, trees can be traversed in different ways.
Generally, there are 2 widely used ways for traversing trees:
- DFS or Depth-First Search
- BFS or Breadth-First Search
What is a Depth-first search?
DFS (Depth-first search) is a technique used for traversing trees or graphs. Here backtracking is used for traversal. In this traversal first, the deepest node is visited and then backtracks to its parent node if no sibling of that node exists
DFS Traversal of a Graph vs Tree:
In the graph, there might be cycles and disconnectivity. Unlike the graph, the tree does not contain a cycle and are always connected. So DFS of a tree is relatively easier. We can simply begin from a node, then traverse its adjacent (or children) without caring about cycles. And if we begin from a single node (root), and traverse this way, it is guaranteed that we traverse the whole tree as there is no dis-connectivity,
Examples:
Input Tree:
Therefore, the Depth First Traversals of this Tree will be:
- Inorder: 4 2 5 1 3
- Preorder: 1 2 4 5 3
- Postorder: 4 5 2 3 1
Below are the Tree traversals through DFS using recursion:
1. Inorder Traversal (Practice):
Follow the below steps to solve the problem:
- Traverse the left subtree, i.e., call Inorder(left-subtree)
- Visit the root
- Traverse the right subtree, i.e., call Inorder(right-subtree)
Below is the implementation of the above algorithm:
C++
// C++ program for different tree 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; struct Node *left, *right; Node(int data) { this->data = data; left = right = NULL; } }; /* Given a binary tree, print its nodes in inorder*/void printInorder(struct Node* node) { if (node == NULL) return; /* first recur on left child */ printInorder(node->left); /* then print the data of node */ cout << node->data << " "; /* now recur on right child */ printInorder(node->right); } /* Driver code*/int main() { struct Node* root = new Node(1); root->left = new Node(2); root->right = new Node(3); root->left->left = new Node(4); root->left->right = new Node(5); cout << "\nInorder traversal of binary tree is \n"; printInorder(root); return 0; } |
C
// C program for different tree traversals #include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data) { struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node); } /* Given a binary tree, print its nodes in inorder*/void printInorder(struct node* node) { if (node == NULL) return; /* first recur on left child */ printInorder(node->left); /* then print the data of node */ printf("%d ", node->data); /* now recur on right child */ printInorder(node->right); } /* Driver code*/int main() { struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); printf("\nInorder traversal of binary tree is \n"); printInorder(root); getchar(); return 0; } |
Java
// Java program for different tree traversals /* Class containing left and right child of current node and key value*/class Node { int key; Node left, right; public Node(int item) { key = item; left = right = null; } } class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes in inorder*/ void printInorder(Node node) { if (node == null) return; /* first recur on left child */ printInorder(node.left); /* then print the data of node */ System.out.print(node.key + " "); /* now recur on right child */ printInorder(node.right); } // Wrappers over above recursive functions void printInorder() { printInorder(root); } // Driver code public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); System.out.println( "\nInorder traversal of binary tree is "); tree.printInorder(); } } |
Python
# Python3 program to for tree traversals # A class that represents an individual node in a # Binary Tree class Node: def __init__(self, key): self.left = None self.right = None self.val = key # A function to do inorder tree traversal def printInorder(root): if root: # First recur on left child printInorder(root.left) # then print the data of node print(root.val), # now recur on right child printInorder(root.right) # Driver code root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) print "\nInorder traversal of binary tree is"printInorder(root) |
C#
// C# program for different tree traversals using System; /* Class containing left and right child of current node and key value*/class Node { public int key; public Node left, right; public Node(int item) { key = item; left = right = null; } } public class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes in inorder*/ void printInorder(Node node) { if (node == null) return; /* first recur on left child */ printInorder(node.left); /* then print the data of node */ Console.Write(node.key + " "); /* now recur on right child */ printInorder(node.right); } // Wrappers over above recursive functions void printInorder() { printInorder(root); } // Driver code public static void Main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); Console.WriteLine( "\nInorder traversal of binary tree is "); tree.printInorder(); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // javascript program for different tree traversals /* Class containing left and right child of current node and key value*/class Node { constructor(val) { this.key = val; this.left = null; this.right = null; } } /* Given a binary tree, print its nodes in inorder */ function printInorder(node) { if (node == null) return; /* first recur on left child */ printInorder(node.left); /* then print the data of node */ document.write(node.key + " "); /* now recur on right child */ printInorder(node.right); } // Driver method var root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); document.write("<br/>Inorder traversal of binary tree is <br/>"); printInorder(root); // This code is contributed by umadevi9616 </script> |
Output
Inorder traversal of binary tree is 4 2 5 1 3
Time Complexity: O(N)
Auxiliary Space: O(log N)
Uses of Inorder traversal:
In the case of binary search trees (BST), Inorder traversal gives nodes in non-decreasing order. To get nodes of BST in non-increasing order, a variation of Inorder traversal where Inorder traversal is reversed can be used
2. Preorder Traversal (Practice):
Follow the below steps to solve the problem:
- Visit the root
- Traverse the left subtree, i.e., call Preorder(left-subtree)
- Traverse the right subtree, i.e., call Preorder(right-subtree)
Below is the implementation of the above algorithm:
C++
// C++ program for different tree 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; struct Node *left, *right; Node(int data) { this->data = data; left = right = NULL; } }; /* Given a binary tree, print its nodes in preorder*/void printPreorder(struct Node* node) { if (node == NULL) return; /* first print data of node */ cout << node->data << " "; /* then recur on left subtree */ printPreorder(node->left); /* now recur on right subtree */ printPreorder(node->right); } /* Driver code*/int main() { struct Node* root = new Node(1); root->left = new Node(2); root->right = new Node(3); root->left->left = new Node(4); root->left->right = new Node(5); cout << "\nPreorder traversal of binary tree is \n"; printPreorder(root); return 0; } |
C
// C program for different tree traversals #include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data) { struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node); } /* Given a binary tree, print its nodes in preorder*/void printPreorder(struct node* node) { if (node == NULL) return; /* first print data of node */ printf("%d ", node->data); /* then recur on left subtree */ printPreorder(node->left); /* now recur on right subtree */ printPreorder(node->right); } /* Driver code*/int main() { struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); printf("\nPreorder traversal of binary tree is \n"); printPreorder(root); getchar(); return 0; } |
Java
// Java program for different tree traversals /* Class containing left and right child of current node and key value*/class Node { int key; Node left, right; public Node(int item) { key = item; left = right = null; } } class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes in preorder*/ void printPreorder(Node node) { if (node == null) return; /* first print data of node */ System.out.print(node.key + " "); /* then recur on left subtree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Wrappers over above recursive functions void printPreorder() { printPreorder(root); } // Driver code public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); System.out.println( "Preorder traversal of binary tree is "); tree.printPreorder(); } } |
Python
# Python3 program to for tree traversals # A class that represents an individual node in a # Binary Tree class Node: def __init__(self, key): self.left = None self.right = None self.val = key # A function to do preorder tree traversal def printPreorder(root): if root: # First print the data of node print(root.val), # Then recur on left child printPreorder(root.left) # Finally recur on right child printPreorder(root.right) # Driver code root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) print "Preorder traversal of binary tree is"printPreorder(root) |
C#
// C# program for different tree traversals using System; /* Class containing left and right child of current node and key value*/public class Node { public int key; public Node left, right; public Node(int item) { key = item; left = right = null; } } public class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes in preorder*/ void printPreorder(Node node) { if (node == null) return; /* first print data of node */ Console.Write(node.key + " "); /* then recur on left subtree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Wrappers over above recursive functions void printPreorder() { printPreorder(root); } // Driver code public static void Main() { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); Console.WriteLine( "Preorder traversal of binary tree is "); tree.printPreorder(); } } /* This code contributed by PrinciRaj1992 */ |
Javascript
<script> // JavaScript implementation of above approach // A class that represents an individual node in a // Binary Tree class Node{ constructor(key){ this.left = null this.right = null this.val = key } } // A function to do preorder tree traversal function printPreorder(root){ if(root){ // First print the data of node document.write(root.val," ") // Then recur on left child printPreorder(root.left) // Finally recur on right child printPreorder(root.right) } } // Driver code let root = new Node(1) root.left = new Node(2) root.right = new Node(3) root.left.left = new Node(4) root.left.right = new Node(5) document.write("Preorder traversal of binary tree is","</br>") printPreorder(root) // This code is contributed by shinjanpatra </script> |
Output
Preorder traversal of binary tree is 1 2 4 5 3
Time Complexity: O(N)
Auxiliary Space: O(log N)
Uses of Preorder:
Preorder traversal is used to create a copy of the tree. Preorder traversal is also used to get prefix expressions of an expression tree.
3. Postorder Traversal (Practice):
Follow the below steps to solve the problem:
- Traverse the left subtree, i.e., call Postorder(left-subtree)
- Traverse the right subtree, i.e., call Postorder(right-subtree)
- Visit the root
Below is the implementation of the above algorithm:
C++
// C program for different tree 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; struct Node *left, *right; Node(int data) { this->data = data; left = right = NULL; } }; /* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */void printPostorder(struct Node* node) { if (node == NULL) return; // first recur on left subtree printPostorder(node->left); // then recur on right subtree printPostorder(node->right); // now deal with the node cout << node->data << " "; } /* Driver code*/int main() { struct Node* root = new Node(1); root->left = new Node(2); root->right = new Node(3); root->left->left = new Node(4); root->left->right = new Node(5); cout << "\nPostorder traversal of binary tree is \n"; printPostorder(root); return 0; } |
C
// C program for different tree traversals #include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data) { struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node); } /* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */void printPostorder(struct node* node) { if (node == NULL) return; // first recur on left subtree printPostorder(node->left); // then recur on right subtree printPostorder(node->right); // now deal with the node printf("%d ", node->data); } /* Driver code*/int main() { struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); printf("\nPostorder traversal of binary tree is \n"); printPostorder(root); getchar(); return 0; } |
Java
// Java program for different tree traversals /* Class containing left and right child of current node and key value*/class Node { int key; Node left, right; public Node(int item) { key = item; left = right = null; } } class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */ void printPostorder(Node node) { if (node == null) return; // first recur on left subtree printPostorder(node.left); // then recur on right subtree printPostorder(node.right); // now deal with the node System.out.print(node.key + " "); } // Wrappers over above recursive functions void printPostorder() { printPostorder(root); } // Driver code public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); System.out.println( "\nPostorder traversal of binary tree is "); tree.printPostorder(); } } |
Python
# Python3 program to for tree traversals # A class that represents an individual node in a # Binary Tree class Node: def __init__(self, key): self.left = None self.right = None self.val = key # A function to do postorder tree traversal def printPostorder(root): if root: # First recur on left child printPostorder(root.left) # the recur on right child printPostorder(root.right) # now print the data of node print(root.val), # Driver code root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) print "\nPostorder traversal of binary tree is"printPostorder(root) |
C#
// C# program for different tree traversals using System; /* Class containing left and right child of current node and key value*/public class Node { public int key; public Node left, right; public Node(int item) { key = item; left = right = null; } } public class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */ void printPostorder(Node node) { if (node == null) return; // first recur on left subtree printPostorder(node.left); // then recur on right subtree printPostorder(node.right); // now deal with the node Console.Write(node.key + " "); } // Wrappers over above recursive functions void printPostorder() { printPostorder(root); } // Driver code public static void Main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); Console.WriteLine( "\nPostorder traversal of binary tree is "); tree.printPostorder(); } } // This code contributed by Rajput-Ji |
Javascript
<script> // javascript program for different tree traversals /* Class containing left and right child of current node and key value*/class Node { constructor(item) { this.key = item; this.left = this.right = null; } } var root; /* * Given a binary tree, print its nodes according to the "bottom-up" postorder * traversal. */ function printPostorder(node) { if (node == null) return; // first recur on left subtree printPostorder(node.left); // then recur on right subtree printPostorder(node.right); // now deal with the node document.write(node.key + " "); } // Driver method root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); document.write("\nPostorder traversal of binary tree is<br/> "); printPostorder(root); // This code contributed by umadevi9616 </script> |
Output
Postorder traversal of binary tree is 4 5 2 3 1
Time Complexity: O(N)
Auxiliary Space: O(log N)
Uses of Postorder:
Postorder traversal is used to delete the tree. Please see the question for the deletion of the tree for details. Postorder traversal is also useful to get the postfix expression of an expression tree
Implementing all traversals using DFS:
C++
// C++ program for different tree 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; struct Node *left, *right; Node(int data) { this->data = data; left = right = NULL; } }; /* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */void printPostorder(struct Node* node) { if (node == NULL) return; // first recur on left subtree printPostorder(node->left); // then recur on right subtree printPostorder(node->right); // now deal with the node cout << node->data << " "; } /* Given a binary tree, print its nodes in inorder*/void printInorder(struct Node* node) { if (node == NULL) return; /* first recur on left child */ printInorder(node->left); /* then print the data of node */ cout << node->data << " "; /* now recur on right child */ printInorder(node->right); } /* Given a binary tree, print its nodes in preorder*/void printPreorder(struct Node* node) { if (node == NULL) return; /* first print data of node */ cout << node->data << " "; /* then recur on left subtree */ printPreorder(node->left); /* now recur on right subtree */ printPreorder(node->right); } /* Driver code*/int main() { struct Node* root = new Node(1); root->left = new Node(2); root->right = new Node(3); root->left->left = new Node(4); root->left->right = new Node(5); cout << "\nPreorder traversal of binary tree is \n"; printPreorder(root); cout << "\nInorder traversal of binary tree is \n"; printInorder(root); cout << "\nPostorder traversal of binary tree is \n"; printPostorder(root); return 0; } |
C
// C program for different tree traversals #include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data) { struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node); } /* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */void printPostorder(struct node* node) { if (node == NULL) return; // first recur on left subtree printPostorder(node->left); // then recur on right subtree printPostorder(node->right); // now deal with the node printf("%d ", node->data); } /* Given a binary tree, print its nodes in inorder*/void printInorder(struct node* node) { if (node == NULL) return; /* first recur on left child */ printInorder(node->left); /* then print the data of node */ printf("%d ", node->data); /* now recur on right child */ printInorder(node->right); } /* Given a binary tree, print its nodes in preorder*/void printPreorder(struct node* node) { if (node == NULL) return; /* first print data of node */ printf("%d ", node->data); /* then recur on left subtree */ printPreorder(node->left); /* now recur on right subtree */ printPreorder(node->right); } /* Driver code*/int main() { struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); printf("\nPreorder traversal of binary tree is \n"); printPreorder(root); printf("\nInorder traversal of binary tree is \n"); printInorder(root); printf("\nPostorder traversal of binary tree is \n"); printPostorder(root); getchar(); return 0; } |
Java
// Java program for different tree traversals /* Class containing left and right child of current node and key value*/class Node { int key; Node left, right; public Node(int item) { key = item; left = right = null; } } class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */ void printPostorder(Node node) { if (node == null) return; // first recur on left subtree printPostorder(node.left); // then recur on right subtree printPostorder(node.right); // now deal with the node System.out.print(node.key + " "); } /* Given a binary tree, print its nodes in inorder*/ void printInorder(Node node) { if (node == null) return; /* first recur on left child */ printInorder(node.left); /* then print the data of node */ System.out.print(node.key + " "); /* now recur on right child */ printInorder(node.right); } /* Given a binary tree, print its nodes in preorder*/ void printPreorder(Node node) { if (node == null) return; /* first print data of node */ System.out.print(node.key + " "); /* then recur on left subtree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Wrappers over above recursive functions void printPostorder() { printPostorder(root); } void printInorder() { printInorder(root); } void printPreorder() { printPreorder(root); } // Driver code public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); System.out.println( "Preorder traversal of binary tree is "); tree.printPreorder(); System.out.println( "\nInorder traversal of binary tree is "); tree.printInorder(); System.out.println( "\nPostorder traversal of binary tree is "); tree.printPostorder(); } } |
Python
# Python3 program to for tree traversals # A class that represents an individual node in a # Binary Tree class Node: def __init__(self, key): self.left = None self.right = None self.val = key # A function to do inorder tree traversal def printInorder(root): if root: # First recur on left child printInorder(root.left) # then print the data of node print(root.val), # now recur on right child printInorder(root.right) # A function to do postorder tree traversal def printPostorder(root): if root: # First recur on left child printPostorder(root.left) # the recur on right child printPostorder(root.right) # now print the data of node print(root.val), # A function to do preorder tree traversal def printPreorder(root): if root: # First print the data of node print(root.val), # Then recur on left child printPreorder(root.left) # Finally recur on right child printPreorder(root.right) # Driver code root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) print "Preorder traversal of binary tree is"printPreorder(root) print "\nInorder traversal of binary tree is"printInorder(root) print "\nPostorder traversal of binary tree is"printPostorder(root) |
C#
// C# program for different Console.Writetree traversals using System; /* Class containing left and right child of current node and key value*/public class Node { public int key; public Node left, right; public Node(int item) { key = item; left = right = null; } } public class BinaryTree { // Root of Binary Tree Node root; BinaryTree() { root = null; } /* Given a binary tree, print its nodes according to the "bottom-up" postorder traversal. */ void printPostorder(Node node) { if (node == null) return; // first recur on left subtree printPostorder(node.left); // then recur on right subtree printPostorder(node.right); // now deal with the node Console.Write(node.key + " "); } /* Given a binary tree, print its nodes in inorder*/ void printInorder(Node node) { if (node == null) return; /* first recur on left child */ printInorder(node.left); /* then print the data of node */ Console.Write(node.key + " "); /* now recur on right child */ printInorder(node.right); } /* Given a binary tree, print its nodes in preorder*/ void printPreorder(Node node) { if (node == null) return; /* first print data of node */ Console.Write(node.key + " "); /* then recur on left subtree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Wrappers over above recursive functions void printPostorder() { printPostorder(root); } void printInorder() { printInorder(root); } void printPreorder() { printPreorder(root); } // Driver code public static void Main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); Console.WriteLine( "Preorder traversal of binary tree is "); tree.printPreorder(); Console.WriteLine( "\nInorder traversal of binary tree is "); tree.printInorder(); Console.WriteLine( "\nPostorder traversal of binary tree is "); tree.printPostorder(); } } // This code has been contributed by 29AjayKumar |
Javascript
<script> // JavaScript program to for tree traversals // A class that represents an individual node in a // Binary Tree class Node{ constructor(key){ this.left = null this.right = null this.val = key } } // A function to do inorder tree traversal function printInorder(root){ if(root){ // First recur on left child printInorder(root.left) // then print the data of node document.write(root.val," ") // now recur on right child printInorder(root.right) } } // A function to do postorder tree traversal function printPostorder(root){ if(root){ // First recur on left child printPostorder(root.left) // the recur on right child printPostorder(root.right) // now print the data of node document.write(root.val," ") } } // A function to do preorder tree traversal function printPreorder(root){ if(root){ // First print the data of node document.write(root.val," ") // Then recur on left child printPreorder(root.left) // Finally recur on right child printPreorder(root.right) } } // Driver code let root = new Node(1) root.left = new Node(2) root.right = new Node(3) root.left.left = new Node(4) root.left.right = new Node(5) document.write("Preorder traversal of binary tree is","</br>") printPreorder(root) document.write("</br>","Inorder traversal of binary tree is","</br>") printInorder(root) document.write("</br>","Postorder traversal of binary tree is","</br>") printPostorder(root) // This code is contributed by shinjanpatra </script> |
Output
Preorder traversal of binary tree is 1 2 4 5 3 Inorder traversal of binary tree is 4 2 5 1 3 Postorder traversal of binary tree is 4 5 2 3 1
Time Complexity: O(N)
Auxiliary Space: O(log N)
Related Article:
Please see this post for Breadth First Traversal.
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