Path Sum

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.

确认是否有等于指定值的路径和  只要计算根节点到每个叶子的路径值之和即可 采用递归 记录到达每个节点时的和 判断该节点是否为叶子 是则进行判断 不是则继续递归 代码如下：

public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
return pathSum(root,sum,0);
}
public boolean pathSum(TreeNode root,int sum,int cur){
if(root==null)return false;
if(root.left==null&&root.right==null)
return sum==cur+root.val? true:false;
if(root.left==null)return pathSum(root.right,sum,cur+root.val);
else{
if(root.right==null)return pathSum(root.left,sum,cur+root.val);
else return pathSum(root.left,sum,cur+root.val)||pathSum(root.right,sum,cur+root.val);
}
}
}