題目描述

????從上到下打印出二叉樹的每個(gè)節(jié)點(diǎn)，同一層的節(jié)點(diǎn)按照從左到右的順序打印。

思路分析

額，好像沒什么好分析的，這就是一個(gè)廣度遍歷，借用隊(duì)列實(shí)現(xiàn)就可以了。

Java代碼實(shí)現(xiàn)

import java.util.Queue;
import java.util.concurrent.LinkedBlockingDeque;

/**
 * 從上倒下逐層遍歷二叉樹（廣度遍歷）
 * @author Administrator
 * @version 2018/10/12
 */
public class Exe33_PrintTreeFromTop2Down {
    public static void main(String[] args) {
        TreeNode node8=new TreeNode(8);
        TreeNode node6=new TreeNode(6);
        TreeNode node10=new TreeNode(10);
        TreeNode node5=new TreeNode(5);
        TreeNode node7=new TreeNode(7);
        TreeNode node9=new TreeNode(9);
        TreeNode node11=new TreeNode(11);
        
        node8.leftTreeNode=node6;node8.rightTreeNode=node10;
        node6.leftTreeNode=node5;node6.rightTreeNode=node7;
        node10.leftTreeNode=node9;node10.rightTreeNode=node11;
        
        print(node8);
    }
    
    //借用隊(duì)列結(jié)構(gòu)實(shí)現(xiàn)
    public static void print(TreeNode root) {
        if(root==null) return;
        else {
            
            Queue<TreeNode> queue=new LinkedBlockingDeque<TreeNode>();
            queue.add(root);
            TreeNode tempNode=null;
            while(!queue.isEmpty()){
                tempNode=queue.poll();
                if(tempNode.leftTreeNode!=null) 
                    queue.add(tempNode.leftTreeNode);
                if(tempNode.rightTreeNode!=null) 
                    queue.add(tempNode.rightTreeNode);
                System.out.print(tempNode.treeVal+" ");
            }
            
        }
    }
}

class TreeNode{
    int treeVal;
    TreeNode leftTreeNode;
    TreeNode rightTreeNode;
    public TreeNode(int value) {
        this.treeVal=value;
    }
    @Override
    public String toString() {
        return "Node"+treeVal;
    }
}