1127 ZigZagging on a Tree (30 分)

1127 ZigZagging on a Tree (30 分)

 

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.

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

由后序和中序确定二叉树。
然后就随便就能输出来了。

#include <bits/stdc++.h>
using namespace std;
int n;
struct Node
{
    int val;
    Node *left, *right;
}*root;
int inorder[50], post[50];
struct mat{
    int val, ord, pri;
};
bool cmp(mat &a, mat &b){
    return a.ord < b.ord;
}
vector<mat> vt;
vector<int> v;
Node *create(int x, int ll, int rr){
    Node *node = (Node*)malloc(sizeof(Node));
    node->val = post[x];
    node->left = NULL, node->right = NULL;
    int mid = 0;
    for(int i = ll; i <= rr; i++){
        if(inorder[i] == post[x]){
            mid = i;
            break;
        }
    }
    if(ll < mid){
        node->left = create(x-rr+mid-1, ll, mid-1);
    }
    if(rr > mid){
        node->right = create(x-1, mid+1, rr);
    }
    return node;
}

void output(Node *root, int x, int y){
    if(root != NULL){
        output(root->left, x<<1, y+1);
        vt.push_back({root->val, x, y});
        output(root->right, (x<<1)+1, y+1);
    }
}

int main(){
    cin >> n;
    for(int i = 0; i < n; i++){
        cin >> inorder[i];
    }
    for(int i = 0; i < n; i++){
        cin >> post[i];
    }
    root = create(n-1, 0, n-1);
    output(root, 1, 1);
    sort(vt.begin(), vt.end(), cmp);
    for(int i = 0; i < vt.size(); i++){
        if(vt[i].pri%2 == 0){
            if(vt[i].pri != vt[i-1].pri){
                int j = i-1;
                while(vt[j].pri == vt[i-1].pri){
                    printf("%d ", vt[j].val);
                    j--;
                }
            }
            printf("%d%c",vt[i].val, i != n-1?' ':'
');
        }
    }
    int j = vt.size()-1;
    if(vt[j].pri%2 == 1){
        while(vt[j].pri == vt[n-1].pri){
            v.push_back(vt[j].val);
            j--;
        }
        for(int i = 0; i < v.size(); i++){
            printf("%d%c", v[i], i == v.size()-1?'
':' ');
        }
    }
    return 0;
}