Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
代码：
/**
* 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:
bool hasPathSum(TreeNode* root, int sum) {
if(root == NULL) return false;
if( root->val == sum &&root->left == NULL && root->right == NULL ) return true;
return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
}
};
今天我偶然发现了一个神奇的功效（下面看对比图）
我觉得一定有更快的方法，但是想不出来。
结果当我把return语句挪到上面后，神奇的事情发生了。