PAT A1119 Pre- and Post-order Traversals

1119 Pre- and Post-order Traversals

分数 30

作者 CHEN, Yue

单位浙江大学

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

给出前序遍历和后序遍历有可能不能唯一的确定一棵二叉树；
* 由于前序遍历访问节点顺序是：根，左，右；
* 后序遍历访问节点顺序是：左，右，根；
* 所以当一棵树只有左子树或者右子树时，那么这棵树就不是唯一的，因为子树
* 既可以在左子树上，也可以在右子树上，这都符合给出的前序遍历和后序遍历；
* 所以当一棵树（或者子树）的只有单边子树时，那么这棵二叉树就不是唯一的；
*
* 又由于题目要求随意给出一棵符合要求的二叉树，不妨先建立左子树，再建立
* 右子树，即只有单边子树的时候将其连接到左子树上；
*
* 首先找到左子树的根节点值v（preL+1位置处的值），根据后序遍历和v来确定
* 左子树的节点个数；递归的建立左子树以及右子树，由于此处需要用到
* pre[preL + 1]的值，因此当preL == preR 的时候，就需要单独处理一下，
* 防止数组越界；

/*** 给出前序遍历和后序遍历有可能不能唯一的确定一棵二叉树；* 由于前序遍历访问节点顺序是：根，左，右；*  后序遍历访问节点顺序是：左，右，根；* 所以当一棵树只有左子树或者右子树时，那么这棵树就不是唯一的，因为子树* 既可以在左子树上，也可以在右子树上，这都符合给出的前序遍历和后序遍历；* 所以当一棵树（或者子树）的只有单边子树时，那么这棵二叉树就不是唯一的；* * 又由于题目要求随意给出一棵符合要求的二叉树，不妨先建立左子树，再建立* 右子树，即只有单边子树的时候将其连接到左子树上；* * 首先找到左子树的根节点值v（preL+1位置处的值），根据后序遍历和v来确定* 左子树的节点个数；递归的建立左子树以及右子树，由于此处需要用到* pre[preL + 1]的值，因此当preL == preR 的时候，就需要单独处理一下，* 防止数组越界；
*/#include <iostream>
#include <algorithm>using namespace std;typedef struct TNode* Bin;
struct TNode
{int v;Bin l, r;
};
const int N = 35;
int pre[N], beh[N];
bool Unique = true,  spa = false;Bin Create(int preL, int preR, int behL, int behR)
{if(preL > preR) return NULL;//由于此处需要用到//pre[preL + 1]的值，因此当preL == preR 的时候，就需要单独处理一下，// 防止数组越界；if(preL == preR){Bin t = new TNode;t->v = pre[preL];t->l = NULL;t->r = NULL;return t;}int v = pre[preL+1]; //左子树的根节点int k;for(k = behL; k <= behR && beh[k] != v; ++k);int numL = k - behL + 1; //得到左子树的节点个数if(preR - preL == numL) //如果是单子树，那么二叉树不唯一Unique = 0;Bin t = new TNode;t->v = pre[preL];t->l = Create(preL+1, preL+numL, behL, k); //左子树t->r = Create(preL+numL+1, preR, k+1, behR-1); //右子树return t;
}void mid_Traver(Bin T)
{if(T){mid_Traver(T->l);if(spa) cout << ' ';else spa = true;cout << T->v;mid_Traver(T->r);}
}
int main()
{int n;cin >> n;for(int i=0; i<n; ++i)  cin >> pre[i];for(int i=0; i<n; ++i)  cin >> beh[i];Bin Tr = Create(0, n-1, 0, n-1);if(Unique)  puts("Yes");else puts("No");mid_Traver(Tr);puts("");return 0;
}