题目

题目链接：
https://www.nowcoder.com/practice/e280b9b5aabd42c9b36831e522485622

思路

 图，队列
构件图，直接从target出发，扩展到第k层就是答案

Java代码

import java.util.*;

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

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param root TreeNode类
     * @param target int整型
     * @param k int整型
     * @return int整型一维数组
     */
    public ArrayList  distanceKnodes (TreeNode root, int target, int k) {
        //构建图
        TreeNode cur = root;
        Map<Integer, List<Integer>> graph = new HashMap<>();
        //BFS,二叉树的层序遍历
        Queue<TreeNode> q = new LinkedList<>();
        q.add(cur);
        graph.put(cur.val, new ArrayList<>());
        while (!q.isEmpty()) {
            TreeNode pop = q.poll();

            if (pop.left != null) {
                q.add(pop.left);
                graph.get(pop.val).add(pop.left.val);

                graph.put(pop.left.val, new ArrayList<>());
                graph.get(pop.left.val).add(pop.val);
            }

            if (pop.right != null) {
                q.add(pop.right);
                graph.get(pop.val).add(pop.right.val);
                graph.put(pop.right.val, new ArrayList<>());
                graph.get(pop.right.val).add(pop.val);
            }
        }


       //图的BFS
        Queue<Integer> queue = new LinkedList<>();
        queue.add(target);
        Set<Integer> visited = new HashSet<>();
        visited.add(target);
        while (k >= 0) {
            k--;


            int size = queue.size();
            for (int i = 0; i < size ; i++) {
                int pop = queue.poll();

                for (Integer next : graph.get(pop)) {
                    if (visited.contains(next)) continue;
                    queue.add(next);
                    visited.add(next);
                }
            }


            if ( k == 0) break;
        }


        ArrayList<Integer> ans = new ArrayList <>();
        int size = queue.size();
        for (int i = 0; i < size ; i++) {
           ans.add( queue.poll());
        }
        return ans;
    }
}

Go代码

package main

import . "nc_tools"

/*
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param root TreeNode类
 * @param target int整型
 * @param k int整型
 * @return int整型一维数组
 */
func distanceKnodes(root *TreeNode, target int, k int) []int {
	//BFS
	graph := map[int][]int{} //构件图

	//层序遍历二叉树
	q1 := []*TreeNode{}
	q1 = append(q1, root)
	graph[root.Val] = []int{}

	for len(q1) > 0 {
		size := len(q1)
		q1bak := []*TreeNode{}
		for i := 0; i < size; i++ {
			pop := q1[i]
			a := pop.Val
			if pop.Left != nil {
				q1bak = append(q1bak, pop.Left)
				b := pop.Left.Val
				graph[a] = append(graph[a], b)
				graph[b] = []int{a}
			}

			if pop.Right != nil {
				q1bak = append(q1bak, pop.Right)
				c := pop.Right.Val
				graph[a] = append(graph[a], c)
				graph[c] = []int{a}
			}
		}

		q1 = q1bak
	}

	//图的BFS
	q := []int{target}
	visited := map[int]bool{}
	visited[target] = true
	for k >= 0 {
		k--
		size := len(q)
		qbak := []int{}
		for i := 0; i < size; i++ {
			pop := q[i]

			nexts := graph[pop]
			for _, next := range nexts {
				_, ok := visited[next]
				if ok {
					continue
				}

				qbak = append(qbak, next)
				visited[next] = true
			}
		}

		q = qbak
		if k == 0 {
			break
		}
	}

	return q
}

PHP代码

<?php

/*class TreeNode{
    var $val;
    var $left = NULL;
    var $right = NULL;
    function __construct($val){
        $this->val = $val;
    }
}*/

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param root TreeNode类 
 * @param target int整型 
 * @param k int整型 
 * @return int整型一维数组
 */
function distanceKnodes( $root ,  $target ,  $k )
{
      //BFS
    $graph = array(); //构建图
    //二叉树的层序遍历
    $q1 = [$root];
    $graph[$root->val] = array();
    while (count($q1) >0){
        $q1bak=array();
        $size = count($q1);
        for($i=0;$i<$size;$i++){
            $pop = $q1[$i];
            $a = $pop->val;

            if($pop->left!=null){
                $q1bak[count($q1bak)] = $pop->left;
                $b = $pop->left->val;
                $graph[$a][count($graph[$a])] = $b;
                $graph[$b]=[$a];
            }

            if($pop->right!=null){
                $q1bak[count($q1bak)] = $pop->right;
                $c = $pop->right->val;
                $graph[$a][count($graph[$a])] = $c;
                $graph[$c]=[$a];
            }
        }

        $q1=$q1bak;
    }

    //图的BFS
    $q = [$target];
    $visited = [$target=>$target];
    while ($k>=0){
        $k--;
        $size = count($q);
        $qbak = array();

        for($i=0;$i<$size;$i++){
            $nexts = $graph[$q[$i]];

            foreach ($nexts as $next){
                if(isset($visited[$next])) continue;

                $qbak[count($qbak)]  =$next;
                $visited[$next] = $next;
            }
        }
        $q = $qbak;
        if($k ==0) break;
    }

    return $q;
}