剑指offer-题23：从上往下打印二叉树

灰太狼 2022-05-29 08:50 175阅读 0赞

## **题目描述** ##

从上往下打印出二叉树的每个节点，同层节点从左至右打印。

实验平台：[牛客网][Link 1]

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## **解决思路：** ##

![这里写图片描述][Image 1]  
![这里写图片描述][Image 1]  
![这里写图片描述][Image 1]  
![这里写图片描述][Image 1]

java:

import java.util.ArrayList;
    import java.util.LinkedList;
    import java.util.Queue;
    /** public class TreeNode { int val = 0; TreeNode left = null; TreeNode right = null; public TreeNode(int val) { this.val = val; } } */
    public class Solution { 
        public ArrayList<Integer> PrintFromTopToBottom(TreeNode root) {
            ArrayList<Integer> arrayList = new ArrayList<>();
            if (root != null) {
                Queue<TreeNode> queue = new LinkedList<TreeNode>();
                queue.offer(root);
                while (!queue.isEmpty()) {
                    TreeNode tempNode = queue.poll();
                    arrayList.add(tempNode.val);
                    if(tempNode.left != null){
                        queue.offer(tempNode.left);
                    }
                    if(tempNode.right != null){
                        queue.offer(tempNode.right);
                    }
                }
            }
            return arrayList;
        }
    }

python:

# -*- coding:utf-8 -*-
    # class TreeNode:
    # def __init__(self, x):
    # self.val = x
    # self.left = None
    # self.right = None
    class Solution:
        # 返回从上到下每个节点值列表，例：[1,2,3]
        def PrintFromTopToBottom(self, root):
            # write code here
            list = []
            if root is not None:
                queue = []
                queue.append(root)
                while len(queue) > 0:
                    tempNode = queue.pop(0)
                    list.append(tempNode.val)
                    if tempNode.left is not None:
                        queue.append(tempNode.left)
                    if tempNode.right is not None:
                        queue.append(tempNode.right)
            return list

[Link 1]: https://www.nowcoder.com/ta/coding-interviews?page=1
[Image 1]: