530. Minimum Absolute Difference in BST

佛系城

于 2017-07-19 21:50:09 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/fujch3/article/details/75452076

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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。

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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;

    }
};