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.
给定了一颗二叉树和一个整数sum,要求我们查找树中是否存在一条从根到叶子节点的和为sum的路径。我们从根节点开始搜索,用DFS,当遍历到一个节点时,从根节点到当前节点的和正好为sum,这时我们还要查看当前节点是否为叶子节点,如果为叶子节点就返回true,否则继续搜索。代码如下:
/** * 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(sum == root.val && root.left == null && root.right == null) return true; return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val); } }