1119 Pre- and Post-order Traversals

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

题意：

　　给出一棵树的前序和中序遍历的结果，判断这棵树是不是唯一，并输出中序遍历的结果。

思路：

　　我们都知道用前序和中序，后序和中序可以确定一颗二叉树，这是因为在一棵子树中前序的第一个结点（或后序的最后一个结点）为根结点，根据根结点在中序中的序列，可以确定确定左子树和右子树，然后递归求解就可以找到一棵确定的二叉树。按照信息论的说法就是前序和中序（或者后序和中序）中包含整棵树的信息。但是，前序遍历和后序遍历，并没有明确的给出根结点和左右子树的关系。只给出前序遍历和后序遍历，可能存在一个结点既可以做左子树，又可以做右子树的情况。那么，这时候这棵树就是不确定的。当然有些特殊的情况，可以只通过前序和后序就可以确定一颗二叉树，那就是根结点既有左子树又有右子树，那么就不存在上面的那种不确定的情况了。也就是说前序和后序序列中所包含的信息+根结点既有左子树又有右子树这样一个特殊的信息也可以确定一棵二叉树。本题假设假设将不确定的孩子结点都放在有子树上。

Code：

#include <bits/stdc++.h>

using namespace std;

typedef struct Node* node;

struct Node {
    int val;
    node left;
    node right;
    Node(int v) {
        val = v;
        left = NULL;
        right = NULL;
    }
};

vector<int> pre, post;
bool isUnique = true;

node buildTree(int preStart, int preEnd, int postStart, int postEnd) {
    int pos = 0, len1, len2;
    if (preStart > preEnd || postStart > postEnd) return NULL;
    node root = new Node(pre[preStart]);
    for (int i = preStart; i <= preEnd; ++i) {
        if (pre[i] == post[postEnd - 1]) {
            pos = i;
            break;
        }
    }
    len1 = preEnd - pos + 1;    // 右子树的长度
    len2 = pos - preStart - 1;  // 左子树的长度

    if (len1 * len2 == 0) isUnique = false;
    if (len1 < 0 || len2 < 0) return root;

    root->left =
        buildTree(preStart + 1, pos - 1, postStart, postEnd - len1 - 1);
    root->right = buildTree(pos, preEnd, postStart + len2, postEnd - 1);

    return root;
}

bool isFirst = true;
void InorderTravelTree(node root) {
    if (root == NULL) return;
    InorderTravelTree(root->left);
    if (isFirst) {
        cout << root->val;
        isFirst = false;
    } else {
        cout << " " << root->val;
    }
    InorderTravelTree(root->right);
}

int main() {
    int n, t;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> t;
        pre.push_back(t);
    }
    for (int i = 0; i < n; ++i) {
        cin >> t;
        post.push_back(t);
    }
    node root = buildTree(0, n - 1, 0, n - 1);

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

    InorderTravelTree(root);
    cout << endl;

    return 0;
}

每一组测试用例后如果不输出换行符会提示“Presentation Error”。