1020 Tree Traversals (25 分)

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博客围绕二叉树遍历问题展开，已知二叉树所有键为不同正整数，给出后序和中序遍历序列，要求输出对应二叉树的层次遍历序列，还提及可用三个数组解决该问题，level数组大小有待考究。

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1020 Tree Traversals (25 分)

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

题解

三个数组即可解决
level数组大小有待考究

code

#include<iostream>
#include <bits/stdc++.h>
using namespace std;

//N=31至少需要？
//101 the last test段错误
//1001 the last test答案错误
//10001 AC

const int maxn = 10001;
int in[maxn];
int post[maxn];
int level[maxn];

void dfs(int root, int st, int ed, int index)
{
    if(st > ed) return;
    level[index] = post[root];
    int i = st;
    while(st < ed && in[i] != post[root]) ++i;
    dfs(root - 1 - ed + i, st, i - 1, 2*index + 1);
    dfs(root - 1, i + 1, ed, 2*index + 2);
}


int main()
{
    int n;
    for(int i = 0; i < maxn; ++i)
        level[i] = -1;
    scanf("%d",&n);
    for(int i = 0; i < n; ++i){
        scanf("%d",&post[i]);
    }
    for(int i = 0; i < n; ++i){
        scanf("%d",&in[i]);
    }
    dfs(n - 1, 0, n - 1, 0);
    for(int i = 0, cnt = 0; i < maxn; ++i){
        if(level[i] != -1){
            printf("%d",level[i]);
            if(cnt != n - 1) printf(" ");
            if(cnt >= n) break;
            cnt++;
        }
    }
    return 0;
}