标签:一个 null == tor tco NPU val cto back
题意:在BST中寻找两个节点,使它们的和为一个给定值。
/**
* 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> v;
void inorder(TreeNode* root){
if(root == NULL) return;
inorder(root -> left);
v.push_back(root -> val);
inorder(root -> right);
}
bool findTarget(TreeNode* root, int k) {
if(root == NULL) return false;
inorder(root);
int len = v.size();
int head = 0;
int tail = len - 1;
while(head < tail){
int sum = v[head] + v[tail];
if(sum == k) return true;
else if(sum > k) --tail;
else ++head;
}
return false;
}
};
LeetCode 653. Two Sum IV - Input is a BST(在BST中寻找两个节点,使它们的和为一个给定值)
标签:一个 null == tor tco NPU val cto back
原文地址:https://www.cnblogs.com/tyty-Somnuspoppy/p/12571732.html