Verify Preorder Sequence in Binary Search Tree

Verify Preorder Sequence in Binary Search Tree - Problem

Given an array of unique integers preorder, return true if it is the correct preorder traversal sequence of a binary search tree.

In a binary search tree, for any node:

All values in the left subtree are less than the node's value
All values in the right subtree are greater than the node's value

In a preorder traversal, we visit nodes in the order: root → left subtree → right subtree

Input & Output

Example 1 — Valid BST Preorder

$ Input: preorder = [5,2,1,3,6]

› Output: true

💡 Note: This represents a valid BST: root=5, left subtree has 2,1,3 (all < 5), right subtree has 6 (> 5). The preorder traversal follows root→left→right pattern.

Example 2 — Invalid BST Preorder

$ Input: preorder = [5,2,6,1,3]

› Output: false

💡 Note: After visiting 6 (right subtree of 5), we cannot visit 1 which is less than 5. This violates BST property as we're in the right subtree where all values should be > 5.

Example 3 — Single Node

$ Input: preorder = [1]

› Output: true

💡 Note: A single node is always a valid BST, so any single element array is a valid preorder sequence.

Constraints

1 ≤ preorder.length ≤ 10⁴
1 ≤ preorder[i] ≤ 10⁸
All integers in preorder are unique

Visualization

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Asked in

G Google 25 a Amazon 20 f Facebook 15 M Microsoft 12

The key insight is to use a monotonic stack to simulate the path from root to current node. The optimal approach maintains a minimum bound and validates each element against BST constraints. Time: O(n), Space: O(n)

Common Approaches

✓ Monotonic Stack

⏱️ Time: O(n) Space: O(n)

Maintain a stack to simulate the path from root to current node. When we encounter a value greater than stack top, we've moved to a right subtree and need to maintain minimum bound.

Recursive Tree Construction

⏱️ Time: O(n) Space: O(n)

For each position, recursively try to build left and right subtrees with proper value bounds. If we can successfully build the entire tree, the sequence is valid.

Monotonic Stack — Algorithm Steps

Step 1: Initialize stack and minimum bound
Step 2: For each element, check if it violates current minimum
Step 3: Pop stack while current element is greater (moving to right subtree)
Step 4: Update minimum bound and push current element

Visualization

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Step-by-Step Walkthrough

Process Each Element

Check against current minimum bound

Pop Smaller Elements

When moving to right subtree, update minimum

Push Current

Add current element to path

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <stdbool.h>

bool solution(int* preorder, int size) {
    int stack[1000];
    int top = -1;
    int minVal = INT_MIN;
    
    for (int i = 0; i < size; i++) {
        int val = preorder[i];
        if (val < minVal) {
            return false;
        }
        
        while (top >= 0 && val > stack[top]) {
            minVal = stack[top];
            top--;
        }
        
        stack[++top] = val;
    }
    
    return true;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int preorder[1000];
    int size = 0;
    
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        preorder[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    bool result = solution(preorder, size);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Each element is pushed and popped from stack at most once

✓ Linear Growth

Space Complexity

O(n)

Stack can contain all elements in worst case (left-skewed tree)

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Monotonic Stack — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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