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.
For 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.
/**
* 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;
}
//leaf node ,no left and right recurision end condition
if(root->left==NULL && root->right==NULL){
return root->val==sum;
}
//left right exist one,return true
bool left=hasPathSum(root->left,sum-root->val);
bool right=hasPathSum(root->right,sum-root->val);
return left || right;
}
};
