[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:

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

思路：

遍历二叉树，将元素值排序，最小的差肯定出现在相邻两个元素里，遍历数组即可。

 1 void preorderTree(TreeNode* root,vector<int>& vc)
 2 {
 3   if(root ==NULL) return ;
 4   vc.push_back(root->val);
 5   preorderTree(root->left,vc);
 6   preorderTree(root->right,vc);
 7 }
 8  int minDiffInBST(TreeNode* root)
 9  {
10    vector<int>vc;
11    preorderTree(root,vc);
12    sort(vc.begin(),vc.end());
13    int t =INT_MAX;
14    for(int i=0;i<vc.size()-1;i++)
15    {
16      t = min(t,vc[i+1] - vc[i]);
17    }
18    return t;
19  }

原文地址：https://www.cnblogs.com/hellowooorld/p/8445419.html