力扣0113——路径总和II

路径总和II

难度：中等

题目描述

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

示例1

输入： root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]

示例2

输入： root = [1,2,3], targetSum = 5
输出：[]

示例3

输入： root = [1,2], targetSum = 0
输出：[]

题解

和 0112 题的思路一致，都是对targetSum进行减法回溯，不同的是需要记录当前节点的值

想法代码

using System;
using System.Collections.Generic;
using System.Linq;
using System.Runtime.Remoting.Metadata.W3cXsd2001;

public class TreeNode
{
    public int val;
    public TreeNode left;
    public TreeNode right;

    public TreeNode(int val = 0, TreeNode left = null, TreeNode right = null)
    {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

public class Solution
{
    public static void Main(string[] args)
    {
        TreeNode root = new TreeNode
        {
            val = 5,
            left = new TreeNode
            {
                val = 4,
                left = new TreeNode
                {
                    val = 11,
                    left = new TreeNode(7),
                    right = new TreeNode(2)
                }
            },
            right = new TreeNode
            {
                val = 8,
                left = new TreeNode(13),
                right = new TreeNode
                {
                    val = 4,
                    left = new TreeNode(5),
                    right = new TreeNode(1)
                }
            }
        };

        Solution solution = new Solution();
        IList<IList<int>> currans = solution.PathSum(root, 22);
        foreach (var a in currans)
        {
            foreach (var b in a)
            {
                Console.Write(b + " ");
            }

            Console.WriteLine();
        }

        Console.ReadKey();
    }


    IList<IList<int>> ans = new List<IList<int>>();
    IList<int> temp = new List<int>();

    public IList<IList<int>> PathSum(TreeNode root, int targetSum)
    {
        BackTrack(root, targetSum);
        return ans;
    }

    public void BackTrack(TreeNode root, int targetSum)
    {
        if (root == null)
        {
            return;
        }

        temp.Add(root.val);

        if (root.val == targetSum && root.left == null && root.right == null)
        {
            ans.Add(new List<int>(temp));
        }

        BackTrack(root.left, targetSum - root.val);
        BackTrack(root.right, targetSum - root.val);
        temp.RemoveAt(temp.Count - 1);
    }
}