Given an array of numbers, return true if given array can represent preorder traversal of a Binary Search Tree, else return false. Expected time complexity is O(n).
Examples:
Input: pre[] = {2, 4, 3}
Output: true
Given array can represent preorder traversal
of below tree
2
\
4
/
3
Input: pre[] = {2, 4, 1}
Output: false
Given array cannot represent preorder traversal
of a Binary Search Tree.
Input: pre[] = {40, 30, 35, 80, 100}
Output: true
Given array can represent preorder traversal
of below tree
40
/ \
30 80
\ \
35 100
Input: pre[] = {40, 30, 35, 20, 80, 100}
Output: false
Given array cannot represent preorder traversal
of a Binary Search Tree.
A Simple Solution is to do following for every node pre[i] starting from first one.
1) Find the first greater value on right side of current node.
Let the index of this node be j. Return true if following
conditions hold. Else return false
(i) All values after the above found greater value are
greater than current node.
(ii) Recursive calls for the subarrays pre[i+1..j-1] and
pre[j+1..n-1] also return true.
Time Complexity of the above solution is O(n2)
An Efficient Solution can solve this problem in O(n) time. The idea is to use a stack. This problem is similar to Next (or closest) Greater Element problem. Here we find the next greater element and after finding next greater, if we find a smaller element, then return false.
1) Create an empty stack.
2) Initialize root as INT_MIN.
3) Do following for every element pre[i]
a) If pre[i] is smaller than current root, return false.
b) Keep removing elements from stack while pre[i] is greater
then stack top. Make the last removed item as new root (to
be compared next).
At this point, pre[i] is greater than the removed root
(That is why if we see a smaller element in step a), we
return false)
c) push pre[i] to stack (All elements in stack are in decreasing
order)
Below is the implementation of above idea.
C++
// C++ program for an efficient solution to check if// a given array can represent Preorder traversal of// a Binary Search Tree#include<bits/stdc++.h>using namespace std;bool canRepresentBST(int pre[], int n){ // Create an empty stack stack<int> s; // Initialize current root as minimum possible // value int root = INT_MIN; // Traverse given array for (int i=0; i<n; i++) { // If we find a node who is on right side // and smaller than root, return false if (pre[i] < root) return false; // If pre[i] is in right subtree of stack top, // Keep removing items smaller than pre[i] // and make the last removed item as new // root. while (!s.empty() && s.top()<pre[i]) { root = s.top(); s.pop(); } // At this point either stack is empty or // pre[i] is smaller than root, push pre[i] s.push(pre[i]); } return true;}// Driver programint main(){ int pre1[] = {40, 30, 35, 80, 100}; int n = sizeof(pre1)/sizeof(pre1[0]); canRepresentBST(pre1, n)? cout << "true\n": cout << "false\n"; int pre2[] = {40, 30, 35, 20, 80, 100}; n = sizeof(pre2)/sizeof(pre2[0]); canRepresentBST(pre2, n)? cout << "true\n": cout << "false\n"; return 0;} |
Java
// Java program for an efficient solution to check if // a given array can represent Preorder traversal of // a Binary Search Tree import java.util.Stack; class BinarySearchTree { boolean canRepresentBST(int pre[], int n) { // Create an empty stack Stack<Integer> s = new Stack<Integer>(); // Initialize current root as minimum possible // value int root = Integer.MIN_VALUE; // Traverse given array for (int i = 0; i < n; i++) { // If we find a node who is on right side // and smaller than root, return false if (pre[i] < root) { return false; } // If pre[i] is in right subtree of stack top, // Keep removing items smaller than pre[i] // and make the last removed item as new // root. while (!s.empty() && s.peek() < pre[i]) { root = s.peek(); s.pop(); } // At this point either stack is empty or // pre[i] is smaller than root, push pre[i] s.push(pre[i]); } return true; } public static void main(String args[]) { BinarySearchTree bst = new BinarySearchTree(); int[] pre1 = new int[]{40, 30, 35, 80, 100}; int n = pre1.length; if (bst.canRepresentBST(pre1, n) == true) { System.out.println("true"); } else { System.out.println("false"); } int[] pre2 = new int[]{40, 30, 35, 20, 80, 100}; int n1 = pre2.length; if (bst.canRepresentBST(pre2, n) == true) { System.out.println("true"); } else { System.out.println("false"); } } } //This code is contributed by Mayank Jaiswal |
Python3
# Python program for an efficient solution to check if # a given array can represent Preorder traversal of # a Binary Search TreeINT_MIN = -2**32def canRepresentBST(pre): # Create an empty stack s = [] # Initialize current root as minimum possible value root = INT_MIN # Traverse given array for value in pre: #NOTE:value is equal to pre[i] according to the #given algo # If we find a node who is on the right side # and smaller than root, return False if value < root : return False # If value(pre[i]) is in right subtree of stack top, # Keep removing items smaller than value # and make the last removed items as new root while(len(s) > 0 and s[-1] < value) : root = s.pop() # At this point either stack is empty or value # is smaller than root, push value s.append(value) return True# Driver Programpre1 = [40 , 30 , 35 , 80 , 100]print ("true" if canRepresentBST(pre1) == True else "false")pre2 = [40 , 30 , 35 , 20 , 80 , 100]print ("true" if canRepresentBST(pre2) == True else "false")# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
// C# program for an efficient solution // to check if a given array can represent // Preorder traversal of a Binary Search Tree using System;using System.Collections.Generic;class GFG{public virtual bool canRepresentBST(int[] pre, int n){ // Create an empty stack Stack<int> s = new Stack<int>(); // Initialize current root as minimum // possible value int root = int.MinValue; // Traverse given array for (int i = 0; i < n; i++) { // If we find a node who is on right side // and smaller than root, return false if (pre[i] < root) { return false; } // If pre[i] is in right subtree of stack top, // Keep removing items smaller than pre[i] // and make the last removed item as new // root. while (s.Count > 0 && s.Peek() < pre[i]) { root = s.Peek(); s.Pop(); } // At this point either stack is empty or // pre[i] is smaller than root, push pre[i] s.Push(pre[i]); } return true;}// Driver Codepublic static void Main(string[] args){ GFG bst = new GFG(); int[] pre1 = new int[]{40, 30, 35, 80, 100}; int n = pre1.Length; if (bst.canRepresentBST(pre1, n) == true) { Console.WriteLine("true"); } else { Console.WriteLine("false"); } int[] pre2 = new int[]{40, 30, 35, 20, 80, 100}; int n1 = pre2.Length; if (bst.canRepresentBST(pre2, n) == true) { Console.WriteLine("true"); } else { Console.WriteLine("false"); }}}// This code is contributed by Shrikant13 |
Javascript
<script>// Javascript program for an efficient// solution to check if a given array // can represent Preorder traversal of// a Binary Search Treefunction canRepresentBST(pre, n){ // Create an empty stack var s = []; // Initialize current root as minimum possible // value var root = -1000000000; // Traverse given array for(var i = 0; i < n; i++) { // If we find a node who is on right side // and smaller than root, return false if (pre[i] < root) return false; // If pre[i] is in right subtree of stack top, // Keep removing items smaller than pre[i] // and make the last removed item as new // root. while (s.length != 0 && s[s.length - 1] < pre[i]) { root = s[s.length - 1]; s.pop(); } // At this point either stack is empty or // pre[i] is smaller than root, push pre[i] s.push(pre[i]); } return true;}// Driver codevar pre1 = [ 40, 30, 35, 80, 100 ];var n = pre1.length;canRepresentBST(pre1, n) ? document.write("true<br>"): document.write("false<br>");var pre2 = [ 40, 30, 35, 20, 80, 100 ];n = pre2.length;canRepresentBST(pre2, n) ? document.write("true"): document.write("false");// This code is contributed by importantly</script> |
Output
true false
Time Complexity: O(n)
Auxiliary Space: O(n)
Another approach:
We can check if the given preorder traversal is valid or not for a BST without using stack. The idea is to use the similar concept of “Building a BST using narrowing bound algorithm”. We will recursively visit all nodes, but we will not build the nodes. In the end, if the complete array is not traversed, then that means that array can not represent the preorder traversal of any binary tree.
Below is the implementation of the above idea:
C++
// C++ program to illustrate if a given array can represent// a preorder traversal of a BST or not#include <bits/stdc++.h>using namespace std;// We are actually not building the treevoid buildBST_helper(int& preIndex, int n, int pre[], int min, int max){ if (preIndex >= n) return; if (min <= pre[preIndex] && pre[preIndex] <= max) { // build node int rootData = pre[preIndex]; preIndex++; // build left subtree buildBST_helper(preIndex, n, pre, min, rootData); // build right subtree buildBST_helper(preIndex, n, pre, rootData, max); } // else // return NULL;}bool canRepresentBST(int arr[], int N){ // code here int min = INT_MIN, max = INT_MAX; int preIndex = 0; buildBST_helper(preIndex, N, arr, min, max); return preIndex == N;}// Driver Code int main(){ int preorder1[] = { 2, 4, 3 }; /* 2 \ 4 / 3*/ int n1 = sizeof(preorder1) / sizeof(preorder1[0]); if (canRepresentBST(preorder1, n1)) cout << "\npreorder1 can represent BST"; else cout << "\npreorder1 can not represent BST :("; int preorder2[] = { 5, 3, 4, 1, 6, 10 }; int n2 = sizeof(preorder2) / sizeof(preorder2[0]); /* 5 / \ 3 1 \ / \ 4 6 10*/ if (canRepresentBST(preorder2, n2)) cout << "\npreorder2 can represent BST"; else cout << "\npreorder2 can not represent BST :("; int preorder3[] = { 5, 3, 4, 8, 6, 10 }; int n3 = sizeof(preorder3) / sizeof(preorder3[0]); /* 5 / \ 3 8 \ / \ 4 6 10*/ if (canRepresentBST(preorder3, n3)) cout << "\npreorder3 can represent BST"; else cout << "\npreorder3 can not represent BST :("; return 0;}// This code is contributed by Sourashis Mondal |
Java
// Java program to illustrate if a given array can represent// a preorder traversal of a BST or notpublic class Main{ static int preIndex = 0; // We are actually not building the tree static void buildBST_helper(int n, int[] pre, int min, int max) { if (preIndex >= n) return; if (min <= pre[preIndex] && pre[preIndex] <= max) { // build node int rootData = pre[preIndex]; preIndex++; // build left subtree buildBST_helper(n, pre, min, rootData); // build right subtree buildBST_helper(n, pre, rootData, max); } // else // return NULL; } static boolean canRepresentBST(int[] arr, int N) { // code here int min = Integer.MIN_VALUE, max = Integer.MAX_VALUE; buildBST_helper(N, arr, min, max); return preIndex == N; } public static void main(String[] args) { int[] preorder1 = { 2, 4, 3 }; /* 2 \ 4 / 3 */ int n1 = preorder1.length; System.out.println(); if (canRepresentBST(preorder1, n1)) System.out.print("preorder1 can represent BST"); else System.out.print("preorder1 can not represent BST :("); int[] preorder2 = { 5, 3, 4, 1, 6, 10 }; int n2 = preorder2.length; System.out.println(); /* 5 / \ 3 1 \ / \ 4 6 10 */ if (!canRepresentBST(preorder2, n2)) System.out.print("preorder2 can represent BST"); else System.out.print("preorder2 can not represent BST :("); int[] preorder3 = { 5, 3, 4, 8, 6, 10 }; int n3 = preorder3.length; System.out.println(); /* 5 / \ 3 8 \ / \ 4 6 10 */ if (canRepresentBST(preorder3, n3)) System.out.print("preorder3 can represent BST"); else System.out.print("preorder3 can not represent BST :("); }}// This code is contributed by mukesh07. |
Python3
# Python3 program to illustrate if a given array can represent# a preorder traversal of a BST or notimport syspreIndex = 0 # We are actually not building the treedef buildBST_helper(n, pre, Min, Max): global preIndex if (preIndex >= n): return if (Min <= pre[preIndex] and pre[preIndex] <= Max): # build node rootData = pre[preIndex] preIndex+=1 # build left subtree buildBST_helper(n, pre, Min, rootData) # build right subtree buildBST_helper(n, pre, rootData, Max) # else # return NULLdef canRepresentBST(arr, N): global preIndex # code here Min, Max = sys.maxsize, -sys.maxsize buildBST_helper(N, arr, Min, Max) if preIndex == N: return True return False preorder1 = [ 2, 4, 3 ]""" 2 \ 4 / 3"""n1 = len(preorder1)if (not canRepresentBST(preorder1, n1)): print("preorder1 can represent BST");else: print("preorder1 can not represent BST :(")preorder2 = [ 5, 3, 4, 1, 6, 10 ]n2 = len(preorder2)""" 5 / \ 3 1 \ / \ 4 6 10"""if (canRepresentBST(preorder2, n2)): print("preorder2 can represent BST")else: print("preorder2 can not represent BST :(")preorder3 = [ 5, 3, 4, 8, 6, 10 ]n3 = len(preorder3)""" 5 / \ 3 8 \ / \ 4 6 10"""if (not canRepresentBST(preorder3, n3)): print("preorder3 can represent BST")else: print("preorder3 can not represent BST :(") |
C#
// C# program to illustrate if a given array can represent// a preorder traversal of a BST or notusing System;class GFG { static int preIndex = 0; // We are actually not building the tree static void buildBST_helper(int n, int[] pre, int min, int max) { if (preIndex >= n) return; if (min <= pre[preIndex] && pre[preIndex] <= max) { // build node int rootData = pre[preIndex]; preIndex++; // build left subtree buildBST_helper(n, pre, min, rootData); // build right subtree buildBST_helper(n, pre, rootData, max); } // else // return NULL; } static bool canRepresentBST(int[] arr, int N) { // code here int min = Int32.MinValue, max = Int32.MaxValue; buildBST_helper(N, arr, min, max); return preIndex == N; } static void Main() { int[] preorder1 = { 2, 4, 3 }; /* 2 \ 4 / 3 */ int n1 = preorder1.Length; Console.WriteLine(); if (canRepresentBST(preorder1, n1)) Console.Write("preorder1 can represent BST"); else Console.Write("preorder1 can not represent BST :("); int[] preorder2 = { 5, 3, 4, 1, 6, 10 }; int n2 = preorder2.Length; Console.WriteLine(); /* 5 / \ 3 1 \ / \ 4 6 10 */ if (!canRepresentBST(preorder2, n2)) Console.Write("preorder2 can represent BST"); else Console.Write("preorder2 can not represent BST :("); int[] preorder3 = { 5, 3, 4, 8, 6, 10 }; int n3 = preorder3.Length; Console.WriteLine(); /* 5 / \ 3 8 \ / \ 4 6 10 */ if (canRepresentBST(preorder3, n3)) Console.Write("preorder3 can represent BST"); else Console.Write("preorder3 can not represent BST :("); }}// This code is contributed by rameshtravel07. |
Javascript
<script> // Javascript program to illustrate if a given array can represent // a preorder traversal of a BST or not let preIndex = 0; // We are actually not building the tree function buildBST_helper(n, pre, min, max) { if (preIndex >= n) return; if (min <= pre[preIndex] && pre[preIndex] <= max) { // build node let rootData = pre[preIndex]; preIndex++; // build left subtree buildBST_helper(n, pre, min, rootData); // build right subtree buildBST_helper(n, pre, rootData, max); } // else // return NULL; } function canRepresentBST(arr, N) { // code here let min = Number.MIN_VALUE, max = Number.MAX_VALUE; buildBST_helper(N, arr, min, max); return preIndex == N; } let preorder1 = [ 2, 4, 3 ]; /* 2 \ 4 / 3 */ let n1 = preorder1.length; if (canRepresentBST(preorder1, n1)) document.write("</br>" + "preorder1 can represent BST"); else document.write("</br>" + "preorder1 can not represent BST :("); let preorder2 = [ 5, 3, 4, 1, 6, 10 ]; let n2 = preorder2.length; /* 5 / \ 3 1 \ / \ 4 6 10 */ if (!canRepresentBST(preorder2, n2)) document.write("</br>" + "preorder2 can represent BST"); else document.write("</br>" + "preorder2 can not represent BST :("); let preorder3 = [ 5, 3, 4, 8, 6, 10 ]; let n3 = preorder3.length; /* 5 / \ 3 8 \ / \ 4 6 10 */ if (canRepresentBST(preorder3, n3)) document.write("</br>" + "preorder3 can represent BST"); else document.write("</br>" + "preorder3 can not represent BST :(");// This code is contributed by decode2207.</script> |
Output
preorder1 can represent BST preorder2 can not represent BST :( preorder3 can represent BST
Time complexity: O(N)
Auxiliary Space: O(height of the possible binary tree)
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