[leetcode] 783. Minimum Distance Between BST Nodes

[leetcode] 783. Minimum Distance Between BST Nodes

Description

Given a Binary Search Tree (BST) with the root node root, return the minimum difference between the values of any two different nodes in the tree.

Example :

Input: root = [4,2,6,1,3,null,null]Output: 1Explanation:Note that root is a TreeNode object, not an array.The given tree [4,2,6,1,3,null,null] is represented by the following diagram: 4 / \ 2 6 / \ 1 3 while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.

Note:

The size of the BST will be between 2 and 100.The BST is always valid, each node’s value is an integer, and each node’s value is different.

分析

题目的意思是：求出二叉搜索树的最小距离。

中序遍历，方法跟minimum-absolute-difference-in-BST一样。

代码

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: int minDiffInBST(TreeNode* root) { int res=INT_MAX; int pre=-1; solve(root,res,pre); return res; } void solve(TreeNode* root,int& res,int& pre){ if(!root){ return ; } solve(root->left,res,pre); if(pre!=-1) res=min(res,root->val-pre); pre=root->val; solve(root->right,res,pre); }};

参考文献

​​[LeetCode] Minimum Distance Between BST Nodes 二叉搜索树中结点的最小距离​​

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