Minimum Absolute Difference in BST

Description

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:

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

Example 2:

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

Constraints:

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

Note: This question is the same as 783: https://leetcode.com/problems/minimum-distance-between-bst-nodes/

Code

Time Complexity: $O(n)$, Space Complexity: $O(1)$

因為是 binary search tree，所以使用 Binary Tree Inorder Traversal。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int getMinimumDifference(TreeNode* root) {
        stack<TreeNode*> st;
        TreeNode* p = root;
        int prev = -1;
        int res = INT_MAX;
        while(p || !st.empty()) {
            if(p) {
                st.push(p);
                p = p->left;
            } else {
                p = st.top(); 
                if(prev != -1) {
                    res = min(res, abs(p->val - prev));
                }
                prev = p->val;
                st.pop();
                p = p->right;
            }
        }
        return res;
    }
};

Source

• Minimum Absolute Difference in BST - LeetCode