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 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 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 program int main() { int pre1[] = {40, 30, 35, 80, 100}; int n = sizeof (pre1)/ sizeof (pre1[0]); canRepresentBST(pre1, n)? cout << "truen" : cout << "falsen" ; int pre2[] = {40, 30, 35, 20, 80, 100}; n = sizeof (pre2)/ sizeof (pre2[0]); canRepresentBST(pre2, n)? cout << "truen" : cout << "falsen" ; 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 |
Python
# Python program for an efficient solution to check if # a given array can represent Preorder traversal of # a Binary Search Tree INT_MIN = - 2 * * 32 def 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 Program pre1 = [ 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 Code public 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 |
Output:
true false
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