leetcode leetcode 783. Minimum Distance Between BST Nodes

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：bst树的中序遍历是一个升序的数组，那么中序遍历一遍，放到vector中，相邻的做差比较就好了。

/**
 * 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:
    priority_queue<int> q;
    void dfs(TreeNode* root) {
        if (root == nullptr) return;
        q.push(root->val);
        dfs(root->left);
        dfs(root->right);
    }
    int minDiffInBST(TreeNode* root) {
        dfs(root);
        int x = q.top();q.pop();
        int ans = 10000000;
        while(!q.empty()) {
            int y = q.top();q.pop();
            ans = min (x-y, ans);
            x = y;
        }
        return ans;
    }
};