1020. Tree Traversals

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

本题考查树的遍历， 给出后根遍历和中根遍历，可以确定一棵树， 最后使用层次遍历获取LevelOrder，思路如下：
直接上例子， 题目给出后根遍历2 3 1 5 7 6 4， 中根遍历1 2 3 4 5 6 7
后跟遍历可知， 4是整棵树的跟， 再看中跟遍历中4的位置，把整棵树分成1,2,3组成的左子树和5,6, 7组成的右子树， 同理， 后根遍历中也会分成2,3,1 和5,7,6两颗子树， 此时，后根2,3,1对应中根1 2 3， 后根5,7,6对应中跟5,6,7可继续递归求解，但是要注意判断特殊情况：

• 确定根后， 发现根在中根遍历的最右边，那么此书右子树为空
• 确定根后， 发现根在中根遍历的最左边，那么此书左子树为空

#include <iostream>
#include <queue>
using namespace std;

typedef struct
{
    int value;
    int left;
    int right;
} Node;

Node node[32];
int postOrder[32];//后跟遍历
int midOrder[32];//中跟遍历
int N;

//递归函数， l1，r1表示后跟遍历的左界和右界
//l2， r2表示中根遍历的左界和右界
void f(int l1, int r1, int l2, int r2)
{
    int root = postOrder[r1];

    if(l1 == r1)
    {
        node[root].left = node[root].right = -1;
        return;
    }
    //i记录根在midOrder的位置
    int i = 0;
    while(midOrder[i] != root) i++;
    int cntL = i - l2;  //cntL表示i左边有几个元素，用于分隔postOrder

    if(i == l2)//左子树为空情况
    {
        node[root].left = -1;
        node[root].right = postOrder[r1 - 1];
        f(l1 + cntL, r1 - 1, i + 1, r2);
        return;
    }
    if(i == r2)//右子树为空情况
    {
        node[root].right = -1;
        node[root].left = postOrder[r1 - 1];
        f(l1, l1 + cntL - 1, l2, i - 1);
        return;
    }
    //两边都不为空，先求解两根
    f(l1, l1 + cntL - 1, l2, i - 1);
    f(l1 + cntL, r1 - 1, i + 1, r2);
    
    //最后给根赋值
    node[root].left = postOrder[l1 + cntL - 1];
    node[root].right = postOrder[r1 - 1];

}
//层次遍历
void bfs()
{
    int root = postOrder[N - 1];
    queue<int> q;
    q.push(root);
    int first = 1;
    while(!q.empty())
    {
        int v = q.front();
        if(first)
        {
            first = 0;
            cout << v;
        }
        else
        {
            cout << " " << v;
        }
        q.pop();
        if(node[v].left != -1)
            q.push(node[v].left);
        if(node[v].right != -1)
            q.push(node[v].right);
    }
}
int main()
{
    cin >> N;

    for(int i = 0; i < N; i++)
    {
        cin >> postOrder[i];
    }
    for(int i = 0; i < N; i++)
    {
        cin >> midOrder[i];
    }

    f(0, N - 1, 0, N - 1);

    bfs();

    return 0;
}