Given the root of a Binary Search Tree (BST), return the minimum absolute difference between the values of any two different nodes in the tree.
Example 1:
Input: root = [4,2,6,1,3]
Output: 1
Example 2:
Input: root = [1,0,48,null,null,12,49]
Output: 1
Constraints:
The number of nodes in the tree is in the range [2, 104].
0 <= Node.val <= 105
Note: This question is the same as 783: https://leetcode.com/problems/minimum-distance-between-bst-nodes/
中序遍历+双指针
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int res = INT_MAX;
TreeNode* pre = nullptr;
//中序遍历,比较两个指针的差值
void traversal(TreeNode* cur){
if(!cur) return;
traversal(cur->left);
if(pre) res = min(res,cur->val - pre->val);
pre = cur;
traversal(cur->right);
}
int getMinimumDifference(TreeNode* root) {
traversal(root);
return res;
}
};