22、剑指offer--从上往下打印二叉树

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。

 

解题思路：本题通过分析，需要采用双端队列对中间节点的信息进行存储。

1）将根节点存入队列中

2）定义间接树的信息pRoot，表示当前遍历的结点信息

3）弹出队列头的结点，然后分别看pRoot的左右孩子是否为空，若不为空则，插入队列尾部

4）循环上述2) 3)操作直至队列为空

#include <iostream>
#include <vector>
#include <deque>
#include <malloc.h>
using namespace std;
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
class Solution {
public:
    //借助辅助的双端队列进行间接存储
    vector<int> PrintFromTopToBottom(TreeNode* root) {
        vector<int> result;
        if(root == NULL)
            return result;
        deque<TreeNode *> dequeTree;
        dequeTree.push_back(root);
        while(!dequeTree.empty())
        {
            TreeNode *pRoot = dequeTree.front();
            result.push_back(pRoot->val);
            dequeTree.pop_front();
            if(pRoot->left)
            {
                dequeTree.push_back(pRoot->left);
            }
            if(pRoot->right)
            {
                dequeTree.push_back(pRoot->right);
            }
        }
    }
};
//先序建立二叉树
TreeNode *CreateBiTree(){
    int ch;

    TreeNode *T;
    cin>>ch;
    if(ch== 0)T=NULL;
    else{
        T = (TreeNode *)malloc(sizeof(TreeNode));
        T->val = ch;
        cout<<"输入左子节点"<<endl;
        T->left = CreateBiTree();
        cout<<"输入右子节点"<<endl;
        T->right = CreateBiTree();
    }
    return T;//返回根节点
}
//先序遍历二叉树
void PreOrderTraverse(TreeNode *T){
    if(T)
    {
       cout<<T->val<<" ";
       PreOrderTraverse(T->left);
       PreOrderTraverse(T->right);
    }
}
int main()
{
    TreeNode *T1;
    cout<<"输入根节点："<<endl;
    T1 = CreateBiTree();//建立
    cout<<"树1的先序遍历为"<<endl;
    PreOrderTraverse(T1);//输出
    cout<<endl;
    cout<<"从上往下打印结果"<<endl;
    vector<int> v;
    Solution s;
    v = s.PrintFromTopToBottom(T1);
    for(int i=0;i<v.size();i++)
    {
        cout<<v[i]<<" ";
    }
    return 0;
}