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.
这道题挺简单的，直接用recursive做就可以了。因为反正每一个node检查的方式都一样。
只要目前的sum减去正在检查的node的value之后等于0，就可以return true。
所以程序还是很简单的。
/**
* 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;
}
sum=sum-root.val;
if(root.left==null&&root.right==null){
if(sum==0){
return true;
}
return false;
}
return hasPathSum(root.left,sum)||hasPathSum(root.right,sum);
}
}