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.
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
我们遍历一条路径结束时,判断这条路径所有节点的和是否等于sum,如果相等的话,我们返回true就可以了,用递归的解法如下.
class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
if(!root)
return false;
if(root->val == sum && root->right == NULL && root->left == NULL)
return true;
return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
}
};