PAT甲级 1020

Tree Traversals

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

---

1. 考点

   • 中序和后序建树
   • 树的层次遍历

2. 题解

   • buildTree 后序-左右中，中序-左中右。由后序最后一个数从中序中找到根结点，则中序的右侧的区间为[i+1,ir]，则可知数目，依此得到后序的区间。继而进行左右子树的分开递归建树。
   • getLevelOrder
   • output

#include<iostream>
#include<queue>
#include<stdio.h>
using namespace std;

const int maxn=32;
int postOrder[maxn],inOrder[maxn],levelOrder[maxn];
typedef struct node{
    int data;
    struct node *lchild,*rchild;
}tree;
queue<tree*> q,p;
tree *buildTree(int pl,int pr,int il,int ir){
    if(pl>pr){
        return NULL;
    }
    int i;
    for(i=il;i<=ir;i++){
        if(postOrder[pr]==inOrder[i]){
            tree *t=(tree*)malloc(sizeof(tree*));
            t->data=postOrder[pr];
            t->lchild=buildTree(pl,pl-il+i-1,il,i-1);
            t->rchild=buildTree(pr-ir+i,pr-1,i+1,ir);
            return t;
        }
    }
}

void getLevelOrder(tree *t){
    int cur=0;
    p.push(t);
    while(!(p.empty())){
        p.swap(q);
        while(!(q.empty())){
            tree *temp=q.front();
            q.pop();
            levelOrder[cur++]=temp->data;

            if(temp->lchild!=NULL)p.push(temp->lchild);
            if(temp->rchild!=NULL)p.push(temp->rchild);
        }
    }
}

void output(int n){
    int i;
    printf("%d",levelOrder[0]);
    for(i=1;i<n;i++){
        printf(" %d",levelOrder[i]);
    }
}

int main(){
    freopen("./in","r",stdin);
    int n;
    scanf("%d",&n);

    int i;
    for(i=0;i<n;i++){
        scanf("%d",postOrder+i);
    }
    for(i=0;i<n;i++){
        scanf("%d",inOrder+i);
    }
    tree *binaryTree;
    binaryTree=buildTree(0,n-1,0,n-1);
    getLevelOrder(binaryTree);
    output(n);
    return 0;
}