Check if a given array can represent Preorder Traversal of Binary Search Tree

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 program
int 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 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

Javascript

<script>

// Javascript program for an efficient
// solution to check if a given array
// can represent Preorder traversal of
// a Binary Search Tree
function 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 code
var 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 tree
void 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 not
public 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 not
import sys

preIndex = 0

# We are actually not building the tree
def 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 NULL

def 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 not
using 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)