127. ZigZagging on a Tree (30)

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本文介绍了一种特殊的二叉树遍历方式——锯齿形层次遍历，即从根节点开始，逐层交替地从左到右和从右到左打印节点值。文章提供了一个实现该遍历方式的示例代码，并详细解释了如何通过构建二叉树并采用两种不同的层次遍历来完成这一任务。

127. ZigZagging on a Tree (30)

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in “zigzagging order” – that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

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 inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.
这里写图片描述
Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. 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:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

127. ZigZagging on a Tree (30)


/*************************************************************
Author : qmeng
MailTo : qmeng1128@163.com
QQ     : 1163306125
Blog   : http://blog.youkuaiyun.com/Mq_Go/
Create : 2018-03-03 16:02:07
Version: 1.0
**************************************************************/
#include <cstdio>
#include <iostream>
#include <queue>
#include <cstdlib>
#include <vector>
using namespace std;
const int MAX = 30;
struct Tree{
    Tree *le;
    Tree *ri;
    int data;
};
bool key = true;
Tree *Tr;
int in[MAX],pos[MAX];
int n;
queue<Tree *> que,queM;
vector<int> v1,v2;
Tree *buildTree(int il,int ir,int pl,int pr){
    if(pl>pr)return NULL;
    int p = il;
    while(in[p] != pos[pr])p++;

    Tree *root = (Tree *) malloc (sizeof(Tree));
    root->data = in[p];
    root->le = buildTree(il,p-1,pl,pr-ir+p-1);
    root->ri = buildTree(p+1,ir,pr-ir+p,pr-1);
    return root;

}
void printLevelTree1(Tree *root){    
    Tree *Tr = NULL;
    que.push(root);
    while(!que.empty()){
        Tree *Tr = que.front();
        que.pop();
        if(Tr==NULL)continue;
        v1.push_back(Tr->data);
        //printf("%d ",Tr->data);
        que.push(Tr->le);
        que.push(Tr->ri);
    }
}
void printLevelTree2(Tree *root){    
    Tree *Tr = NULL;
    que.push(root);
    while(!que.empty()){
        Tree *Tr = que.front();
        que.pop();
        if(Tr==NULL)continue;
        v2.push_back(Tr->data);
        //printf("%d ",Tr->data);
        que.push(Tr->ri);
        que.push(Tr->le);
    }
}
/*
1 11 5 12 17 8 20 15
1 5 11 8 17 12 15 20
*/
void print(int k){
    if(k>=n)return;
    int p = k;
    if(key == true){
        while(v1[p]!=v2[k]){
            printf(" %d",v1[p]);
            p++;
        }
        printf(" %d",v1[p]);
        key = false;
    }else{
        while(v2[p]!=v1[k]){
            printf(" %d",v2[p]);
            p++;
        }
        printf(" %d",v2[p]);
        key = true;
    }

    print(++p);
}
int main(){
    int i;
    scanf("%d",&n);
    for( i = 0 ; i < n ; i++)scanf("%d",&in[i]);
    for( i = 0 ; i < n ; i++)scanf("%d",&pos[i]);
    Tree *root =buildTree(0,n-1,0,n-1);
    printLevelTree1(root);
    printLevelTree2(root);
    printf("%d",v1[0]);
    print(1);

    return 0;
}