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 思路：递归所有路径，如果到达叶子节点，并且此时sum=0，则返回true;
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root==null)
return flase; //一开始如果传进来空引用，则返回false;
sum-=root.val;
if (root.left==null&&&root.right==null){ //如果已经到了叶子
if (sum==0) //如果此时sum=0，则返回true
return true; //否则返回false
else
return false;
}
//同时递归左右子树
return hasPathSum(root.left,sum)||hasPathSum(root.right,sum);
}
}