1020 Tree Traversals

中后序建树+层序遍历

(今天很emo,没怎么做事,来灌个水

题面

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

简单的中后序建树+简单的遍历,随便灌个水,代码如下(

// 中后序建树+层序遍历,一遍过
#include <cstdio>
#include <iostream>
#include <queue>
#define MAXN 100001
using namespace std;
struct node
{
    int v;
    node *l;
    node *r;
};
int in[MAXN];
int mappost[MAXN]; // 节点值对下标的反映射,空间换时间方便找
int post[MAXN];

int n;
void init()
{
    for (int i = 1; i <= n; i++)
    {
        mappost[post[i]] = i;
    }
}
node *findroot(int inl, int inr, int postl, int postr) //包含头尾,递归分治
{
    if (inl > inr)
    {
        return NULL;
    }
    node *c = new node();
    c->v = post[postr];
    int i, j;
    for (i = inl; i <= inr; i++)
    {
        if (in[i] == post[postr])
        {
            break;
        }
    }
    int m = -1;
    for (j = inl; j <= i - 1; j++)
    {
        m = max(m, mappost[in[j]]);
    }
    c->l = findroot(inl, i - 1, postl, m);
    c->r = findroot(i + 1, inr, m + 1, postr - 1);
    return c;
}
node *root;
void LOT() //层序遍历
{
    queue<node *> q;
    q.push(root);
    while (!q.empty())
    {
        node *c = q.front();
        q.pop();

        cout << c->v;
        if (c->l != NULL)
        {
            q.push(c->l);
        }
        if (c->r != NULL)
        {
            q.push(c->r);
        }
        if (!q.empty())
        {
            cout << " ";
        }
    }
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> post[i];
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> in[i];
    }
    init();
    root = findroot(1, n, 1, n);
    LOT();
    return 0;
}

灌水完毕,明天出去玩。

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