LeetCode //C - 530. Minimum Absolute Difference in BST

530. Minimum Absolute Difference in BST

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.
 

Example 1:

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

Example 2:

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

Constraints:
  • The number of nodes in the tree is in the range [ 2 , 1 0 4 2, 10^4 2,104].
  • 0 < = N o d e . v a l < = 1 0 5 0 <= Node.val <= 10^5 0<=Node.val<=105

From: LeetCode
Link: 530. Minimum Absolute Difference in BST


Solution:

Ideas:

The strategy to solve this problem is:

  1. Perform an in-order traversal of the BST.
  2. As we traverse, compare the difference between the current node and the previous node.
  3. Keep track of the minimum difference encountered.
Code:
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int prev = -1; 
int minDiff = INT_MAX;

void inorder(struct TreeNode* root) {
    
    
    if (!root) return;

    inorder(root->left);

    // If prev is initialized (i.e., not -1), compute the difference
    if (prev != -1) {
    
    
        int diff = root->val - prev;
        if (diff < minDiff) {
    
    
            minDiff = diff;
        }
    }
    prev = root->val;

    inorder(root->right);
}

int getMinimumDifference(struct TreeNode* root) {
    
    
    // Initialize global variables for each new call
    prev = -1;
    minDiff = INT_MAX;

    inorder(root);
    return minDiff;
}

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转载自blog.csdn.net/navicheung/article/details/132962562