leetcode113. Path Sum II

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   1return[   [5,4,11,2],   [5,8,4,5]]

解法

DFS，判断如果是根节点，并且符合条件，则加入列表中，向上一层要把新添加的元素remove掉。

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> pathSum(TreeNode root, int sum) {        List<List<Integer>> ret = new ArrayList<>();        List<Integer> list = new ArrayList<>();        if (root == null) {            return ret;        }        helper(root, sum, list, ret);        return ret;    }    public void helper(TreeNode root, int sum, List<Integer> list, List<List<Integer>> ret) {        if (root == null) {            return;        }        list.add(root.val);        if (root.left == null && root.right == null && sum - root.val == 0) {            ret.add(new ArrayList<Integer>(list));            list.remove(list.size() - 1);            return;        } else {            helper(root.left, sum - root.val, list, ret);            helper(root.right, sum - root.val, list, ret);        }        list.remove(list.size() - 1);    }}