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783. 二叉搜索树节点最小距离

给你一个二叉搜索树的根节点 root ,返回 树中任意两不同节点值之间的最小差值

差值是一个正数,其数值等于两值之差的绝对值。

 

示例 1:

输入:root = [4,2,6,1,3]
输出:1

示例 2:

输入:root = [1,0,48,null,null,12,49]
输出:1

 

提示:

  • 树中节点的数目范围是 [2, 100]
  • 0 <= Node.val <= 105

 

注意:本题与 530:https://leetcode.cn/problems/minimum-absolute-difference-in-bst/ 相同

相似题目

二叉树的中序遍历

原站题解

去查看

/**
* 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 minDiffInBST(TreeNode* root) {
}
};
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golang 解法, 执行用时: 0 ms, 内存消耗: 2.3 MB, 提交时间: 2021-07-13 17:33:29

/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func minDiffInBST(root *TreeNode) int {
ans, pre := math.MaxInt64, -1
var dfs func(*TreeNode)
dfs = func(node *TreeNode) {
if node == nil {
return
}
dfs(node.Left)
if pre != -1 && node.Val -pre < ans {
ans = node.Val - pre
}
pre = node.Val
dfs(node.Right)
}
dfs(root)
return ans
}

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