Leetcode 113. Path Sum II 路径和2 解题报告

1 解题思想

这道题和Leetcode 112是同宗的，112只要求输出是否存在一个Case，而这题要求输出所有Case的路径。
方法基本是一样他，遍历到底的时候保存一下路径就好

Leetcode 112. Path Sum 路径和 解题报告

2 原题

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]
]


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3 AC解

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */


 /**
  * 和原来的不一样，这题要完全遍历，使用track跟踪当前的进度，进入dfs时压进去，出dfs时推出，当时叶节点且和正好为0是，克隆添加到结果中。
  * */
public class Solution {
    List<List<Integer>> list;
    LinkedList<Integer> track;
    public void dfs(TreeNode root,int sum){
        if(root==null)
            return;
        sum-=root.val;
        track.add(root.val);
        if(sum==0 && root.left==null && root.right==null)
            list.add((LinkedList<Integer>)track.clone());
        dfs(root.left,sum);
        dfs(root.right,sum);
        track.remove(track.size()-1);

    }
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        this.list=new ArrayList<List<Integer>>();
        this.track=new LinkedList<Integer>();
        dfs(root,sum);
        return this.list;
    }
}