[LeetCode] 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.

找出BST中两个节点的最小值。

1、利用中序遍历将书中节点值按顺序放入数组中。

2、找出数组中相邻元素间差的绝对值的最小值即可。

/**
 * 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:
    vector<int> inVec;
    int minDiffInBST(TreeNode* root) {
        inOrder(root);
        int res = INT_MAX;
        for (int i = 1; i < inVec.size(); i++) {
            res = min(res, abs(inVec[i] - inVec[i - 1]));
        }
        return res;
    }
    
    void inOrder(TreeNode* root) {
        if (root == nullptr)
            return;
        inOrder(root->left);
        inVec.push_back(root->val);
        inOrder(root->right);
    }
};
// 6 ms