[Leetcode] 783. Minimum Distance Between BST Nodes 解题报告

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本文介绍了一种在二叉搜索树中寻找两不同节点间最小值差的方法。通过中序遍历二叉搜索树，记录遍历过程中相邻节点值之间的最小差值。此算法空间复杂度为O(1)，时间复杂度为O(n)。

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题目：

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: 1
Explanation:
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.

思路：

我们只需要对BST进行中序遍历，找出遍历中相邻节点之间差异的最小值即可。算法的空间复杂度是O(1)，时间复杂度是O(n)。

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) {
        if (root->left) {
            minDiffInBST(root->left);
        }
        if (first) {
            first = !first;
            pre_value = root->val;
        }
        else {
            ret = min(ret, root->val - pre_value);
            pre_value = root->val;
        }
        if (root->right) {
            minDiffInBST(root->right);
        }
        return ret;
    }
private:
    int ret = INT_MAX, pre_value = -1;
    bool first = true;
};