給你一個(gè)二叉樹，請(qǐng)你返回其按 層序遍歷 得到的節(jié)點(diǎn)值。 （即逐層地，從左到右訪問所有節(jié)點(diǎn)）

https://leetcode-cn.com/problems/binary-tree-level-order-traversal/

示例1：

二叉樹：[3,9,20,null,null,15,7],

3
/ \
9 20
/ \
15 7

[
[3],
[9,20],
[15,7]
]

Java解法

思路：

• 層序邊歷，通過棧緩存當(dāng)前數(shù)據(jù)，因?yàn)橄冗M(jìn)后出，所以先存儲(chǔ)右節(jié)點(diǎn)，
• 優(yōu)化使用隊(duì)列來存儲(chǔ)，但效率沒有太多提升

package sj.shimmer.algorithm.m3_2021;

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
import java.util.Queue;
import java.util.Stack;

import sj.shimmer.algorithm.TreeNode;

/**
 * Created by SJ on 2021/3/18.
 */

class D52 {
    public static void main(String[] args) {
        System.out.println(levelOrder(TreeNode.getInstance(new Integer[]{3, 9, 20, null, null, 15, 7})));
        System.out.println(levelOrder(TreeNode.getInstance(new Integer[]{1, 2, 3, 4, null, null, 5})));
        System.out.println(levelOrder(TreeNode.getInstance(new Integer[]{3, 9, 20, null, null, 15, 7})));
    }
    public static List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> results = new ArrayList<>();
        Queue<TreeNode> queue = new ArrayDeque<>();
        if (root != null) {
            queue.add(root);
        }
        while (!queue.isEmpty()) {
            List<Integer> tempList = new ArrayList<>();
            Queue<TreeNode> temp = new ArrayDeque<>();
            while (!queue.isEmpty()) {
                TreeNode poll = queue.poll();
                tempList.add(poll.val);
                if (poll.left != null) {
                    temp.add(poll.left);
                }
                if (poll.right != null) {
                    temp.add(poll.right);
                }
            }
            results.add(tempList);
            queue = temp;
        }
        return results;
    }
    public static List<List<Integer>> levelOrder1(TreeNode root) {
        List<List<Integer>> results = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        if (root != null) {
            stack.add(root);
        }
        while (!stack.isEmpty()){
            List<Integer> tempList = new ArrayList<>();
            Stack<TreeNode> temp = new Stack<>();
            while (!stack.isEmpty()) {
                TreeNode pop = stack.pop();
                if (pop != null) {
                    tempList.add(pop.val);
                    temp.add(pop.left);
                    temp.add(pop.right);
                }
            }
            if (tempList.size()!=0) {
                results.add(tempList);
                while (!temp.isEmpty()) {
                    stack.add(temp.pop());
                }
            }
        }
        return results;
    }
}

image

image

官方解

https://leetcode-cn.com/problems/binary-tree-level-order-traversal/solution/er-cha-shu-de-ceng-xu-bian-li-by-leetcode-solution/

1. 廣度優(yōu)先搜索

   相較我的處理，優(yōu)化了臨時(shí)存儲(chǔ)

   • 時(shí)間復(fù)雜度：O(n)

   • 空間復(fù)雜度：O(n)