题目描述
给定一个二叉树和一个值sum，判断是否有从根节点到叶子节点的节点值之和等于sum的路径，
例如：
给出如下的二叉树，sum=22，
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
返回true，因为存在一条路径5->4->11->2的节点值之和为22
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 andsum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.
例如：
给出如下的二叉树，sum=22，
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
返回true，因为存在一条路径5->4->11->2的节点值之和为22
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 andsum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.
示例2
输出
true
/**
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
class Solution {
public:
/**
*
* @param root TreeNode类
* @param sum int整型
* @return bool布尔型
*/
bool hasPathSum(TreeNode* root, int sum) {
// write code here
if (root==NULL){
return false;
}
if(root->left ==NULL && root->right==NULL && sum-root->val==0)
return true;
return hasPathSum(root->left, sum-root->val)||hasPathSum(root->right, sum-root->val);
}
};