[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。

 void preorderTree(TreeNode* root,vector<int>& vc)
 {
   if(root ==NULL) return ;
   vc.push_back(root->val);
   preorderTree(root->left,vc);
   preorderTree(root->right,vc);
 }
  int minDiffInBST(TreeNode* root)
  {
    vector<int>vc;
    preorderTree(root,vc);
    sort(vc.begin(),vc.end());
    int t =INT_MAX;
    for(int i=;i<vc.size()-;i++)
    {
      t = min(t,vc[i+] - vc[i]);
    }
    return t;
  }