Tree Traversals (25)

最新推荐文章于 2023-08-19 19:24:19 发布

s136424

最新推荐文章于 2023-08-19 19:24:19 发布

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本文链接：https://blog.youkuaiyun.com/s136424/article/details/78290604

本文介绍了一种根据给定的后序遍历和中序遍历序列构建二叉树，并输出该树的层次遍历序列的方法。通过递归创建二叉树节点，最终实现了从输入序列到层次遍历输出的有效转换。

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题目描述

Suppose that all the keys in a binary tree are distinct positive integers.  Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

输入描述:

Each input file contains one test case.  For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree.  The second line gives the postorder sequence and the third line gives the inorder sequence.  All the numbers in a line are separated by a space.

输出描述:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree.  All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

输入例子:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

输出例子:

4 1 6 3 5 7 2

我的代码：

#include<iostream>
#include<queue>
using namespace std;
struct Node
{
    int data;
    Node *lchild,*rchild;
};
Node *creat(int h[],int z[],int n)
{
    Node *t;
	int i;
    if(n==0) return NULL;
    t=new Node;
    t->data=h[n-1];
    for(i=0;i<n;i++)
    {
        if(z[i]==h[n-1]) break;
    }
    t->lchild=creat(h,z,i);
    t->rchild=creat(h+i,z+i+1,n-i-1);
    return t;
}
void ceng(Node *t)
{
    queue<Node*>q;
    int a[101],i,k=0;
    if(t)
    {
        q.push(t);
        while(!q.empty())
        {
            t=q.front();
            q.pop();
            a[k++]=t->data;
            if(t->lchild) q.push(t->lchild);
            if(t->rchild) q.push(t->rchild);
        }
    }
    for(i=0;i<k;i++)
	{
		if(i==0) cout<<a[i];
		else cout<<" "<<a[i];
	}
}
int main()
{
    int n,i,b[101],c[101];
    cin>>n;
    for(i=0;i<n;i++) cin>>b[i];
    for(i=0;i<n;i++) cin>>c[i];
    Node *r=creat(b,c,n);
    ceng(r);
    return 0;
}