1119.Pre- and Post-order Traversals

题目描述

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

考点

1.树的三种遍历方式；
2.前序后序转中序。

思路

1.根据特点求解
前序的输出顺序为根左右，后序为左右根，因此只要前序的子序列的第一个值等于后序子序列的最后一个值，这个值就是根结点的值。可以通过这个特点递归建树。
2.中序遍历
3.如何确定是否唯一？
不唯一的情况就是某棵子树的根结点只有一个孩子结点，那么这个孩子结点可以是左孩子，也可以是右孩子。反之，如果某棵子树有左右孩子结点那么它一定是唯一可确定的，即最终前序和后序两个数组一定可以被分割成[a, b]和[a, b]的形式，即prel=prer且postl=postr。如果prel>prer，那么不可唯一确定。

代码

#include 
#include 
#include 
using namespace std;
vector pre, post;
queue in;
int tree[30][2], root, isuniq=1;
void inorder(int &index, int prel,int prer, int postl, int postr) {
    if (prel > prer) {
        isuniq = 0;return;
    }
    index = prel;
    if (prel == prer) return;
    int e = postr;
    while (pre[prel + 1] != post[e]) e--;
    inorder(tree[index][0], prel + 1, prel + e - postl + 1, postl, e);
    inorder(tree[index][1], prel + e - postl + 2, prer, e + 1, postr-1);
}
void inorder(int r) {
    if (tree[r][0] != 0) inorder(tree[r][0]);
    in.push(pre[r]);
    if (tree[r][1] != 0) inorder(tree[r][1]);
}
int main() {
    int n, i, t;
    cin >> n;
    pre.resize(n); post.resize(n);
    for (i = 0; i < n; i++) cin >> pre[i];
    for (i = 0; i < n; i++) cin >> post[i];
    inorder(root, 0, n - 1, 0, n - 1);
    cout << (isuniq == 1 ? "Yes" : "No") << endl;
    inorder(0);
    while (!in.empty()) {
        cout << in.front() << (in.size() == 1 ? "\n" : " ");
        in.pop();
    }
    return 0;
}