【题目】
Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.
For example:
Given the below binary tree and sum
,
= 22
5 / 4 8 / / 11 13 4 / \ / 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
【解析】
DFS,递归实现。
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { List<List<Integer>> res = new ArrayList<List<Integer>>(); public List<List<Integer>> pathSum(TreeNode root, int sum) { if (root == null) return res; List<Integer> list = new ArrayList<Integer>(); list.add(root.val); dfs(root, sum - root.val, list); return res; } public void dfs(TreeNode root, int sum, List<Integer> list) { if (root.left == null && root.right == null && sum == 0) { res.add(list); return; } if (root.left != null) { List<Integer> leftList = new ArrayList<Integer>(list); leftList.add(root.left.val); dfs(root.left, sum - root.left.val, leftList); } if (root.right != null) { List<Integer> rightList = new ArrayList<Integer>(list); rightList.add(root.right.val); dfs(root.right, sum - root.right.val, rightList); } } }
时间: 2024-11-09 00:38:36