Minimum Absolute Difference in BST

Minimum Absolute Difference in BST - Problem

Given the root of a Binary Search Tree (BST), return the minimum absolute difference between the values of any two different nodes in the tree.

The absolute difference between two values a and b is |a - b|, which is always non-negative.

A BST is a binary tree where for every node, all values in the left subtree are smaller and all values in the right subtree are larger than the node's value.

Input & Output

Example 1 — Basic BST

$ Input: root = [4,2,6,1,3]

› Output: 1

💡 Note: In-order traversal gives [1,2,3,4,6]. Adjacent differences are: |2-1|=1, |3-2|=1, |4-3|=1, |6-4|=2. Minimum is 1.

Example 2 — Larger Differences

$ Input: root = [1,0,48,null,null,12,49]

› Output: 1

💡 Note: In-order traversal: [0,1,12,48,49]. Differences: |1-0|=1, |12-1|=11, |48-12|=36, |49-48|=1. Minimum is 1.

Example 3 — Single Node

$ Input: root = [1]

› Output: 0

💡 Note: Only one node, so no pairs exist. Return 0 by convention.

Constraints

The number of nodes in the tree is in the range [2, 10⁴]
0 ≤ Node.val ≤ 10⁵

Visualization

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Asked in

G Google 35 A Amazon 28 M Microsoft 22 🍎 Apple 18

The key insight is that in a BST, the minimum difference always occurs between adjacent nodes in the sorted order. The optimal approach uses in-order DFS traversal to visit nodes in ascending order and compare each node with the previously visited one. Time: O(n), Space: O(h).

Common Approaches

✓ Brute Force - All Pairs Comparison

⏱️ Time: O(n²) Space: O(n)

Collect all node values into an array, then check all possible pairs to find the minimum absolute difference. This approach is straightforward but inefficient.

DFS In-Order Traversal

⏱️ Time: O(n) Space: O(h)

Since a BST's in-order traversal gives nodes in ascending order, we can find the minimum difference by comparing each node with the previously visited node during traversal.

Brute Force - All Pairs Comparison — Algorithm Steps

Traverse tree and collect all values
Compare every pair of values
Track minimum difference found

Visualization

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Step-by-Step Walkthrough

Collect Values

Traverse tree and store all node values

Compare Pairs

Check difference between every pair of values

Find Minimum

Track the smallest absolute difference found

Code -

solution.c — C

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

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

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

void collect_values(struct TreeNode* node, int* values, int* count) {
    if (!node) return;
    values[(*count)++] = node->val;
    collect_values(node->left, values, count);
    collect_values(node->right, values, count);
}

int solution(struct TreeNode* root) {
    if (!root) return 0;
    
    int values[10000];
    int count = 0;
    collect_values(root, values, &count);
    
    int min_diff = INT_MAX;
    for (int i = 0; i < count; i++) {
        for (int j = i + 1; j < count; j++) {
            int diff = abs(values[i] - values[j]);
            if (diff < min_diff) {
                min_diff = diff;
            }
        }
    }
    
    return min_diff == INT_MAX ? 0 : min_diff;
}

void parseArray(const char* str, int* arr, char** null_flags, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        
        if (strncmp(p, "null", 4) == 0) {
            null_flags[*size] = malloc(5);
            strcpy(null_flags[*size], "null");
            arr[*size] = 0;
            p += 4;
        } else {
            null_flags[*size] = NULL;
            arr[*size] = (int)strtol(p, (char**)&p, 10);
        }
        (*size)++;
    }
}

struct TreeNode* buildTree(int* arr, char** null_flags, int size, int index) {
    if (index >= size || null_flags[index] != NULL) return NULL;
    struct TreeNode* node = createNode(arr[index]);
    node->left = buildTree(arr, null_flags, size, 2 * index + 1);
    node->right = buildTree(arr, null_flags, size, 2 * index + 2);
    return node;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int arr[1000];
    char* null_flags[1000];
    int size;
    parseArray(line, arr, null_flags, &size);
    
    struct TreeNode* root = buildTree(arr, null_flags, size, 0);
    int result = solution(root);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops to compare all n*(n-1)/2 pairs of nodes

⚠ Quadratic Growth

Space Complexity

O(n)

Array to store all n node values plus recursion stack

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Ln 1, Col 1

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Related Problems

Common Approaches

Brute Force - All Pairs Comparison — Algorithm Steps

Visualization

Code -

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