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.
给定一棵树和一个数，判断这个数是不是从根节点到叶子节点的和。
递归实现很简单。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) { if( root == null)
return false;
if( root.left == null && root.right == null)
return sum==root.val; return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val); }
}