PAT——1119 Pre- and Post-order Traversals (30 分)（給定先序和後序序列，判斷能否構建一顆唯一的樹）

1119 Pre- and Post-order Traversals (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, or preorder and inorder traversal sequences. However, if only the postorder and preorder traversal sequences are given, the corresponding tree may no longer be unique.

Now given a pair of postorder and preorder traversal sequences, you are supposed to output the corresponding inorder traversal sequence of the tree. If the tree is not unique, simply output any one of them.

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 preorder 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, first printf in a line Yes if the tree is unique, or No if not. Then print in the next line the inorder traversal sequence of the corresponding binary tree. If the solution is not unique, any answer would do. It is guaranteed that at least one solution exists. 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 1:

7
1 2 3 4 6 7 5
2 6 7 4 5 3 1

Sample Output 1:

Yes
2 1 6 4 7 3 5

Sample Input 2:

4
1 2 3 4
2 4 3 1

Sample Output 2:

No
2 1 3 4

思路：關鍵在於父節點是否有一個孩子：

• 如果只有一個孩子，構建樹不唯一
• 如果有兩個孩子或沒有孩子，建樹唯一

#include <iostream>
#include <stdio.h>
#include <queue>
#include <vector>
#include <string.h>
using namespace std;

int n;
int pre[31], post[31];
vector<int> in;
bool uniquee = true;

typedef struct binTree
{
    int data;
    binTree *l_child;
    binTree *r_child;
    binTree()
    {
        l_child = r_child = NULL;
    }
}binTree;

void input()
{
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        cin >> pre[i];
    }
    for(int i = 0; i < n; i++)
        cin >> post[i];
}

binTree *create_one(int preL, int preR, int postL, int postR)
{
    if(preL > preR || postL > postR)
    {
        return NULL;
    }

    int data = pre[preL];
    binTree *root = new binTree();
    root->data = data;
    if(preL == preR)
    {
        root->l_child = NULL;
        root->r_child = NULL;
        return root;
    }
    if(preL + 1 == preR)
    {
        uniquee = false;
        root->l_child = NULL;
        root->r_child = create_one(preL+1, preR, postL, postR-1);
        return root;
    }
    /*查找左子樹的結束座標*/
    int left_data = pre[preL+1];//左根元素
    int i;
    for(i = postL; i <= postR-1; i++)
    {
        if(post[i] == left_data)
        {
            break;
        }
    }
    int left_len = i - postL + 1;
    root->l_child = create_one(preL+1, preL+left_len,postL, postL+left_len-1);
    root->r_child = create_one(preL+left_len+1, preR, postL+left_len, postR-1);
    return root;
}

void inorder(binTree *root)
{
    if(root == NULL)
        return;
    inorder(root->l_child);
    in.push_back(root->data);
    inorder(root->r_child);
}

int main()
{
    /*1119 Pre- and Post-order Traversals (30 分)*/
    #ifndef ONLINE_JUDGE
        freopen("test.txt", "r", stdin);
    #endif // ONLINE_JUDGE

    input();

    binTree *root = create_one(0, n-1, 0, n-1);

    if(uniquee)
        cout << "Yes"<<endl;
    else
        cout << "No"<<endl;

    inorder(root);
    cout << in[0];
    for(int i = 1; i < n; i++)
        cout <<" "<< in[i];
    cout <<endl;
    return 0;
}