Leetcode-113 Path Sum II

问题描述

地址：https://leetcode.com/problems/path-sum-ii/#/description
描述：
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]]

翻译：
给出一个二叉树和一个sum值，找出所有从根节点到叶子节点的路径，这些路径需要满足：路径上所有节点的和=sum

问题解析

设置两个list innerList 和 outerList
（1）从根节点向下遍历，一边遍历一边将节点值加和，并向innerList 中追加
（2）当遇到叶子节点的时候判断加和是否与sum相等

• 如果相等，则将innerList 添加到outerList中，同时去掉innerList中最后一个元素；

• 否则继续向innerList中追加

  解析代码

public List<List<Integer>> pathSum(TreeNode root, int sum) {        List<List<Integer>> resultList = new ArrayList<List<Integer>>();        List<Integer> innerList = new ArrayList<Integer>();        pathSum(resultList, root, 0, innerList, sum);        return resultList;    }    private void pathSum(List<List<Integer>> outerList, TreeNode root, int count, List<Integer> innerList, int sum) {        if(root == null){            return ;        }        count += root.val;        if(count == sum && root.left == null && root.right == null){            innerList.add(root.val);            outerList.add(new ArrayList<Integer>(innerList));            innerList.remove(innerList.size()-1);            return;        }        innerList.add(root.val);        pathSum(outerList, root.left, count, innerList, sum);        pathSum(outerList, root.right, count, innerList, sum);        innerList.remove(innerList.size()-1);    }