530. Minimum Absolute Difference in BST

最新推荐文章于 2022-08-14 16:27:15 发布

原创最新推荐文章于 2022-08-14 16:27:15 发布 · 218 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

108 篇文章

订阅专栏

本文介绍了一种求解二叉搜索树中任意两节点间最小绝对差的方法。通过深度优先搜索遍历所有节点并将值存入数组，随后排序数组并计算相邻元素间的差值，从而找到最小绝对差。

题目来源【Leetcode】

Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.

Example:

Input:
…..1
…….\
…….. 3
……./
…..2
Output:
1

Explanation:
The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).
Note: There are at least two nodes in this BST.

先遍历将每个点的值存在数组里，然后排序将相邻的数相减取绝对值再找出最小的

void dfs(TreeNode*root,vector<int>& re){
    if(!root) return;
     re.push_back(root->val);
    dfs(root->left,re);
    dfs(root->right,re);
}
class Solution {
public:
    int getMinimumDifference(TreeNode* root) {
        vector<int>re;
        dfs(root,re);
        sort(re.begin(),re.end());
        int mi = INT_MAX;
        for(int i = 1; i < re.size(); i++ ){
            mi = min(mi,abs(re[i]-re[i-1]));
        }
        return mi;

    }
};