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.
给定一棵二叉树，问能不能有一条路径（从根节点走到叶子结点），相加的和等于sum值。
递归。
bool hasPathSum(TreeNode* root, int sum) {
if(!root) return false;
if(root->val == sum && !root->left && !root->right) return true;
return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
}
