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:
The size of the BST will be between 2 and 100.
The BST is always valid, each node's value is an integer, and each node's value is different.
思路：bst树的中序遍历是一个升序的数组，那么中序遍历一遍，放到vector中，相邻的做差比较就好了。
/**
* 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:
priority_queue<int> q;
void dfs(TreeNode* root) {
if (root == nullptr) return;
q.push(root->val);
dfs(root->left);
dfs(root->right);
}
int minDiffInBST(TreeNode* root) {
dfs(root);
int x = q.top();q.pop();
int ans = 10000000;
while(!q.empty()) {
int y = q.top();q.pop();
ans = min (x-y, ans);
x = y;
}
return ans;
}
};