LeetCode 863 二叉树中所有距离为K的节点
给定一个二叉树(具有根结点 root
), 一个目标结点 target
,和一个整数值 K
。
返回到目标结点 target
距离为 K
的所有结点的值的列表。 答案可以以任何顺序返回。
示例 1:
输入:root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2 输出:[7,4,1] 解释: 所求结点为与目标结点(值为 5)距离为 2 的结点, 值分别为 7,4,以及 1![]()
注意,输入的 "root" 和 "target" 实际上是树上的结点。
上面的输入仅仅是对这些对象进行了序列化描述。
package com.example.lettcode.tree;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
/**
* @Class DistanceK
* @Description 863 二叉树中所有距离为K的节点
* @Author
* @Date 2021/7/28
**/
public class DistanceK {
static class TreeNode {
int val;
TreeNode left;
TreeNode right;
public TreeNode(int val) {
this.val = val;
this.left = null;
this.right = null;
}
}
static Map<TreeNode, TreeNode> treeNodeParentMap = new HashMap<>();
static List<Integer> ans = new ArrayList<>();
/**
* (1) 先找出每个节点的父节点
* (2) 然后使用bfs,从目标节点出发,找到距离为K的节点
*
* @param root
* @param target
* @param k
* @return
*/
public static List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
if (root == null || target == null) return new ArrayList<>();
ans = new ArrayList<>();
treeNodeParentMap = new HashMap<>();
// 构建节点与其父节点的映射关系
findParents(root);
// 从目标节点出发,找到距离为K的节点
findTargetNode(target, null, 0, k);
return ans;
}
/**
* 寻找每个节点的父节点
*
* @param treeNode
*/
private static void findParents(TreeNode treeNode) {
if (treeNode == null) return;
if (treeNode.left != null) {
treeNodeParentMap.put(treeNode.left, treeNode);
findParents(treeNode.left);
}
if (treeNode.right != null) {
treeNodeParentMap.put(treeNode.right, treeNode);
findParents(treeNode.right);
}
}
/**
* 从目标节点开始寻找距离为K的节点
*
* @param node
* @param from
* @param depth
* @param K
*/
private static void findTargetNode(TreeNode node, TreeNode from, int depth, int K) {
if (node == null) return;
// 该节点与目标节点的距离为K
if (depth == K) {
ans.add(node.val);
return;
}
// 分别从左右子树开始寻找距离为K的节点
if (node.left != from) {
findTargetNode(node.left, node, depth + 1, K);
}
if (node.right != from) {
findTargetNode(node.right, node, depth + 1, K);
}
// 从父节点开始寻找距离为K的节点,就是从父节点的另一子节点开始寻找
if (treeNodeParentMap.get(node) != from) {
findTargetNode(treeNodeParentMap.get(node), node, depth + 1, K);
}
}
}
测试用例
public static void main(String[] args) {
// root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2
TreeNode root = new TreeNode(3);
TreeNode treeNode02 = new TreeNode(5);
TreeNode treeNode03 = new TreeNode(1);
root.left = treeNode02;
root.right = treeNode03;
TreeNode treeNode04 = new TreeNode(6);
TreeNode treeNode05 = new TreeNode(2);
treeNode02.left = treeNode04;
treeNode02.right = treeNode05;
TreeNode treeNode06 = new TreeNode(0);
TreeNode treeNode07 = new TreeNode(8);
treeNode03.left = treeNode06;
treeNode03.right = treeNode07;
TreeNode treeNode08 = new TreeNode(7);
TreeNode treeNode09 = new TreeNode(4);
treeNode05.left = treeNode08;
treeNode05.right = treeNode09;
TreeNode target = treeNode02;
int K = 2;
List<Integer> integerList = DistanceK.distanceK(root, target, K);
System.out.print("DistanceK demo01 result : [");
for (Integer integer : integerList) {
System.out.print(", " + integer);
}
System.out.println("]");
root = new TreeNode(1);
target = root;
K = 3;
integerList = DistanceK.distanceK(root, target, K);
System.out.print("DistanceK demo02 result : [");
for (Integer integer : integerList) {
System.out.print(", " + integer);
}
System.out.println("]");
}
运行结果
DistanceK demo01 result : [, 7, 4, 1]
DistanceK demo02 result : []