PAT1020

　　最近在刷PAT题，将自己的解题过程和代码写在blog上。

 Tree Traversals (25)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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.

Input Specification:

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.

Output Specification:

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.

Sample Input:

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

Sample Output:

4 1 6 3 5 7 2

   这题自己想了挺久，根据后序排列和中序排列的数组来确定一棵二叉树 。解题思路：根据后序排列可以确定树的根，根据中序排列可以比对根节点确定左右子树。然后因为后序排列安照 左-》右-》根的顺序遍历，而中序排列按照左->根->右的顺序，可以

确定后序排列中的从左到右直到达到根节点左边的肯定为左子树；即不断确定根节点，左右子树来完善这颗树。

　　当树行成后需要按层级打印，这里注意不是按照先序遍历的方式打印，这时需要用到队列，将根元素先入队，再将左右子树节点入队，根据节点不为空的原则进行出队操作，一直到队列为空；

//
//  main.cpp
//  1020
//
//  Created by apple on 14-2-13.
//  Copyright (c) 2014年 apple. All rights reserved.
//

#include <iostream>
#include <queue>
using namespace std;
#define  MAX 31
int inorder[MAX];
int postorder[MAX];
typedef struct Node{
    int value;
    struct Node *lchild;
    struct Node *rchlid;
}Node;

void LevelTravel(Node *root);
Node *createtree(int istart,int iend,int pstart,int pend)
{
    int i,j;
    Node *node = new Node();
    
    node->lchild = NULL;
    node->rchlid = NULL;
    node->value = postorder[pend];
    if (istart>iend || pstart>pend)
    {
        return NULL;
    }
    

    for(i = istart,j = pstart; i < iend && j < pend; i++,j++)
    {
        if (inorder[i] == postorder[pend])
        {
            break;
        }
    }
    
    
    node->lchild = createtree(istart,i-1,pstart,j-1);
    node->rchlid = createtree(i+1,iend,j,pend-1);
    
    
    return node;
}



int main()
{
    int i;
    int num;
    cin>>num;
    for (i = 0; i<num; i++)
    {
        cin>>postorder[i];
    }
    
    for (i = 0; i<num; i++)
    {
        cin>>inorder[i];
    }
    
    Node *root = NULL;
    root = createtree(0,num-1,0,num-1);
    
   LevelTravel(root);
    
    return 0;
}

void LevelTravel(Node *root)
{
    queue<Node*>q;
    if (root!=NULL)
    {
        printf("%d",root->value);
        if(root->lchild!=NULL)q.push(root->lchild);
        if(root->rchlid!=NULL)q.push(root->rchlid);
    }
    else
        return;
    while(!q.empty())
    {
        Node *node = q.front();
        
        if (node == NULL)
        {
            continue;
        }
                q.pop();
        printf(" %d",node->value);
    
        if(node->lchild!=NULL) q.push(node->lchild);
        
        if(node->rchlid!=NULL) q.push(node->rchlid);
        
    }
    printf("
");
    
}