Description
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,
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
Solution
DFS
注意镊叁，null節(jié)點(diǎn)的val是undefined抖剿，不能認(rèn)為是0敞咧！遞歸的終止是null或者left node撤嫩。
/**
* 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 false;
if (root.left == null && root.right == null) return root.val == sum;
return hasPathSum(root.left, sum - root.val)
|| hasPathSum(root.right, sum - root.val);
}
}
