Construct Binary Search Tree from Preorder Traversal

Construct Binary Search Tree from Preorder Traversal - Problem

Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root.

It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.

A binary search tree is a binary tree where for every node, any descendant of Node.left has a value strictly less than Node.val, and any descendant of Node.right has a value strictly greater than Node.val.

A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left, then traverses Node.right.

Input & Output

Example 1 — Basic BST

$ Input: preorder = [8,5,1,7,10,12]

› Output: [8,5,10,1,7,null,12]

💡 Note: Preorder traversal visits root first (8), then left subtree (5,1,7), then right subtree (10,12). Reconstructed tree maintains BST property.

Example 2 — Single Node

$ Input: preorder = [1]

› Output: [1]

💡 Note: Single node forms a tree with just the root.

Example 3 — Right Skewed Tree

$ Input: preorder = [1,3,5,7]

› Output: [1,null,3,null,5,null,7]

💡 Note: Each node only has a right child, forming a right-skewed tree.

Constraints

1 ≤ preorder.length ≤ 100
1 ≤ preorder[i] ≤ 10⁸
The values of preorder are unique

Visualization

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Asked in

G Google 35 a Amazon 28 M Microsoft 22 f Facebook 18

The key insight is to use the BST property to determine node placement. The stack approach is optimal - maintain a monotonic decreasing stack of potential parents and attach each new node appropriately. Time: O(n), Space: O(n)

Common Approaches

✓ Dp 1d

⏱️ Time: N/A Space: N/A

Monotonic Stack Approach

⏱️ Time: O(n) Space: O(n)

Maintain a stack of potential parents. For each node, pop elements that are smaller (they can't be ancestors), then attach to the last popped element or top of stack.

Recursive with Bounds

⏱️ Time: O(n²) Space: O(n)

For each node in preorder, recursively determine if it belongs in left or right subtree based on BST property constraints.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

typedef struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
} TreeNode;

TreeNode* createNode(int val) {
    TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
    node->val = val;
    node->left = NULL;
    node->right = NULL;
    return node;
}

int index_global;

TreeNode* buildBST(int* preorder, int size, int minVal, int maxVal) {
    if (index_global >= size) return NULL;
    
    int val = preorder[index_global];
    if (val < minVal || val > maxVal) return NULL;
    
    index_global++;
    TreeNode* root = createNode(val);
    
    root->left = buildBST(preorder, size, minVal, val);
    root->right = buildBST(preorder, size, val, maxVal);
    
    return root;
}

TreeNode* bstFromPreorder(int* preorder, int preorderSize) {
    index_global = 0;
    return buildBST(preorder, preorderSize, INT_MIN, INT_MAX);
}

void printLevelOrder(TreeNode* root) {
    if (!root) {
        printf("[]");
        return;
    }
    
    TreeNode* queue[1000];
    int front = 0, rear = 0;
    queue[rear++] = root;
    
    printf("[");
    int first = 1;
    
    while (front < rear) {
        TreeNode* node = queue[front++];
        
        if (!first) printf(",");
        first = 0;
        
        if (node) {
            printf("%d", node->val);
            queue[rear++] = node->left;
            queue[rear++] = node->right;
        } else {
            printf("null");
        }
    }
    
    // Remove trailing nulls
    while (rear > 0 && !queue[rear-1]) {
        rear--;
    }
    
    printf("]");
}

void printArray(TreeNode* root) {
    if (!root) {
        printf("[]");
        return;
    }
    
    TreeNode* queue[1000];
    int front = 0, rear = 0;
    queue[rear++] = root;
    
    printf("[");
    int first = 1;
    
    while (front < rear) {
        TreeNode* node = queue[front++];
        
        if (!first) printf(",");
        first = 0;
        
        printf("%d", node->val);
        
        if (node->left) queue[rear++] = node->left;
        if (node->right) queue[rear++] = node->right;
    }
    
    printf("]");
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse array from format [1,2,3,4]
    int preorder[1000];
    int size = 0;
    
    char* ptr = line;
    while (*ptr && size < 1000) {
        if (*ptr == '-' || (*ptr >= '0' && *ptr <= '9')) {
            int num = 0;
            int sign = 1;
            if (*ptr == '-') {
                sign = -1;
                ptr++;
            }
            while (*ptr >= '0' && *ptr <= '9') {
                num = num * 10 + (*ptr - '0');
                ptr++;
            }
            preorder[size++] = sign * num;
        } else {
            ptr++;
        }
    }
    
    if (size == 0) {
        printf("[]");
        return 0;
    }
    
    TreeNode* root = bstFromPreorder(preorder, size);
    printArray(root);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

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