标签:oid dfs 示例 面试 pat int path roo bsp
输入一棵二叉树和一个整数,打印出二叉树中节点值的和为输入整数的所有路径。从树的根节点开始往下一直到叶节点所经过的节点形成一条路径。
示例:
给定如下二叉树,以及目标和 sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
返回:
[
[5,4,11,2],
[5,8,4,5]
]
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { List<List<Integer>> ans = new ArrayList<>(); public List<List<Integer>> pathSum(TreeNode root, int sum) { dfs(root, sum, 0, new ArrayList<>()); return ans; } public void dfs(TreeNode root, int sum, int value, List<Integer> list) { if (root == null) return; list.add(root.val); if (sum == value + root.val && root.left == null && root.right == null) { ans.add(new ArrayList<>(list)); } dfs (root.left, sum, value + root.val, list); dfs (root.right, sum, value + root.val,list); list.remove(list.size() - 1); } }
标签:oid dfs 示例 面试 pat int path roo bsp
原文地址:https://www.cnblogs.com/ziytong/p/13163318.html