1127 ZigZagging on a Tree

题目：

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 (

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

题目大意：给出一个树的中序和后序遍历结果，求它的Z字型层序遍历，也就是偶数层从右往左，奇数层从左往右遍历

思路：

用一个变量保存每个结点的深度，结果按深度层层输出，正向或反向输出每一层的所有结点。

代码：（满分）

#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std;
int n;
int in[35], post[35];
bool flag = false, dir = false;
struct Node{
    int w, level; 
    Node *lchild, *rchild;
};
vector<int> level[35];
Node *build(int inl, int inr, int postl, int postr){
    if(postl > postr){
        return NULL;
    }
    Node *root =  new Node;
    root -> w = post[postr];
   
    int k;
    for(int i = inl; i<=inr;i++){
        if(in[i] == post[postr]){
            k = i;
            break;
        }
    }
    root -> lchild = build(inl, k - 1, postl, postl + k - inl - 1);
    root -> rchild = build(k + 1, inr, postl + k - inl, postr - 1);
    return root;
}
void levelOrder(Node *root){
    queue<Node*> q;
    q.push(root);
    while(!q.empty()){
        Node *tmp = q.front();
        q.pop();
        Node·*l·=·tmp·->·lchild;
        Node *r = tmp -> rchild;
        if(l != NULL){
            l -> level = tmp -> level + 1;
            q.push(l);
            level[l -> level].push_back(l -> w);
        }
        if(r != NULL){
            r -> level = tmp -> level + 1;
            q.push(r);
            level[r -> level].push_back(r -> w);
        }
     }

}
int main(){
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%d", &in[i]);
    }
    for(int i = 0; i < n; i++){
        scanf("%d", &post[i]);
    }
    Node *root = build(0, n - 1, 0, n - 1);
    root -> level = 0;
    levelOrder(root);
    printf("%d", root->w);
    for(int i = 0; i < 35; i++){
        if(dir){
            for(int j = 0; j < level[i].size(); j++){
               
                printf(" %d", level[i][j]);
            }
            dir = false;
        }else{
            for(int j = level[i].size() - 1; j >=0; j--){
                printf(" %d", level[i][j]);
            }
            dir = true;
        }
    }
    return 0;
}