/**
* 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->left == NULL && root->right == NULL && sum != root->val){
return false;
}
else if(root->left == NULL && root->right == NULL && sum == root->val)
return true;
return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
}
};
遇见树能用递归就用递归,okay!