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.
解题思路:
递归拆分子问题:
1.root为空 返回fasle
2.root是叶子节点 且root.val==sum 返回true
3.递归拆分 左节点或者右节点满足 都返回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 false; if(root.left==null&&root.right==null&&root.val==sum) return true; else return hasPathSum(root.left, sum-root.val)||hasPathSum(root.right, sum-root.val); } }