#数据结构与算法学习笔记#剑指Offer58:把二叉树打印成多行(Java、C/C++)

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2019.2.27     《剑指Offer》从零单刷个人笔记整理(66题全)目录传送门​​​​​​​

其实这道题就是层序遍历二叉树,之前已经写过了:#数据结构与算法学习笔记#剑指Offer21:从上往下打印二叉树/层次遍历二叉树(Java、C/C++)

相当于复习一下,没什么好讲的。难一点的是上一道题,按之字型打印二叉树:#数据结构与算法学习笔记#剑指Offer57:按之字形顺序打印二叉树(Java、C/C++)


题目描述

从上到下按层打印二叉树,同一层结点从左至右输出。每一层输出一行。


Java实现:

/**
 * 
 * @author ChopinXBP
 * 从上到下按层打印二叉树,同一层结点从左至右输出。每一层输出一行。
 * 
 */

import java.util.ArrayList;
import java.util.LinkedList;

public class PrintTree_59 {
	
	public static class TreeNode {
	    int val = 0;
	    TreeNode left = null;
	    TreeNode right = null;

	    public TreeNode(int val) {
	        this.val = val;
	    }
	}
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		TreeNode pHead = new TreeNode(0);
		TreeNode p1= new TreeNode(1);
		TreeNode p2= new TreeNode(2);
		TreeNode p3= new TreeNode(3);
		TreeNode p4= new TreeNode(4);
		TreeNode p5= new TreeNode(5);
		TreeNode p6= new TreeNode(6);
		pHead.left = p1;
		pHead.right = p2;
		p1.left = p3;
		p1.right = p4;
		p2.left = p5;
		p2.right = p6;
		
		ArrayList<ArrayList<Integer>> result = Print(pHead);
		for(int i = 0; i < result.size(); i++) {
			for(int j = 0; j < result.get(i).size(); j++) {
				System.out.print(result.get(i).get(j) + " ");
			}
			System.out.println(" ");
		}
	}
	
	public static ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {       
    	ArrayList<ArrayList<Integer>> result = new ArrayList<>();
    	if(pRoot == null)return result;
    	
    	LinkedList<TreeNode> tree = new LinkedList<>(); 
    	tree.add(pRoot);
    	
    	while(!tree.isEmpty()) {
    		TreeNode pNode;
    		ArrayList<Integer> list = new ArrayList<>();   		
    		int num = tree.size();   			
			for (int i = 0; i < num; i++) {
				pNode = tree.pollFirst();
				if (pNode.left != null) {
					tree.add(pNode.left);
				}
				if (pNode.right != null) {
					tree.add(pNode.right);
				}
				list.add(pNode.val);
			}
    		if(!list.isEmpty()) result.add(list);
    	}
    	
    	return result;
    }

}

C++实现示例:

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
        vector<vector<int> > Print(TreeNode* pRoot) {
            vector<vector<int> > vv;
            vector<int> ve;
            queue<TreeNode*> que;
            queue<int> dque;
            int depth=0;
            que.push(pRoot);
            dque.push(depth);
            TreeNode *node;
            while(!que.empty()){
                node=que.front();
                if(depth!=dque.front()){
                    vv.push_back(ve);
                    ve.clear();
                    depth=dque.front();
                }
                que.pop();
                dque.pop();
                if(node){
                    que.push(node->left);
                    que.push(node->right);
                    dque.push(depth+1);
                    dque.push(depth+1);
                    ve.push_back(node->val);
                }
            }
            return vv;
        }
     
};

#Coding一小时,Copying一秒钟。留个言点个赞呗,谢谢你#

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转载自blog.csdn.net/qq_20304723/article/details/87980981