783. Minimum Distance of the Binary Search Tree node

Given the root node of a binary search treerootReturns the minimum deviation of any two vertices in the tree.

Example:

Input:Root = [4, 2, 6, 1, 3, null, null]Output:1Explanation:Note that root is a tree Node object instead of an array. The given tree [4/2, 3, null, null] can be expressed as: 6/\ 1 3. The smallest difference is 1, which is the difference between node 1 and node 2, it is also the difference between node 3 and node 2.

Note:

1. The size range of the binary tree is2To100.
2. Binary Trees are always valid. The values of each node are integers and are not repeated.

       int minDiffInBST(TreeNode* root) {    vector<int > res;    inorder(root, res);    int Min = 65535;    for (int i = 0; i<res.size() - 1; i++){        Min = min(Min, res[i + 1] - res[i]);    }    return Min;}TreeNode * inorder(TreeNode* root, vector<int >&res){    if (root==NULL)        return NULL;    inorder(root->left, res);    res.push_back(root->val);    inorder(root->right, res);    return root;}

    

783. Minimum Distance of the Binary Search Tree node