标签：equal example null for each new equals roo space
Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.
Note: A leaf is a node with no children.
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] ]
time: O(n), space: O(height)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public List<List<Integer>> pathSum(TreeNode root, int sum) { List<List<Integer>> res = new ArrayList<>(); List<Integer> list = new ArrayList<>(); pathSum(root, sum, list, res); return res; } public void pathSum(TreeNode root, int sum, List<Integer> list, List<List<Integer>> res) { if(root == null) { return; } if(root.left == null && root.right == null) { if(sum == root.val) { list.add(root.val); res.add(new ArrayList<>(list)); list.remove(list.size() - 1); } return; } list.add(root.val); pathSum(root.left, sum - root.val, list, res); pathSum(root.right, sum - root.val, list, res); list.remove(list.size() - 1); } }
标签：equal example null for each new equals roo space
原文地址：https://www.cnblogs.com/fatttcat/p/10198743.html
