PAT (Advanced Level) Practice 1119 Pre- and Post-order Traversals （30 分）

• 编程题

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

题目大意：

给出一棵树的前序和后序遍历，求该树是否唯一，并输出其中一种中序遍历的结果。

解题思路：

结果不唯一，因为如果某个节点只有左子树或右子树，无法区分是左边还是右边。

方法：遇到这种情况默认左边或者默认右边。

如何判断某个节点只有左子树或者右子树：

如果后序的根节点左边和前序的根节点右边是同一个点，说明缺一个子树。

注意事项：

如果前序的 start 和 end 是同一个节点，不会进入while循环，也会出现 i=start+1 的情况。需要把这种情况剔除。

源代码：

#include<iostream>
#include<stdlib.h>
#include<vector>
using namespace std;
struct node
{
	int data;
	int left = -1, right = -1;
};
vector<node>v;
int pre[31], in[31], post[31], flag=0, idx=0;
void to_in(int root, int start, int end)//root=post[end],start=pre[start],end=pre[end]
{
	if (start > end)return;
	int i = start + 1;
	while (i <= end && pre[i] != post[root - 1])i++;
	if (i == start + 1&&start!=end)	flag = 1; 
	to_in(root-(end-i+1)-1,start+1,i-1);
	in[idx++] = post[root];
	to_in(root-1,i,end);
	return;
}
int main()
{
	int n, i, j, k;
	scanf("%d", &n);
	for (i = 0; i < n; i++)scanf("%d", &pre[i]);
	for (i = 0; i < n; i++)scanf("%d", &post[i]);
	to_in(n-1, 0, n-1);
	if (flag == 1)printf("No\n");
	else printf("Yes\n");
	for (i = 0; i < n; i++)
		printf("%d%c", in[i], i == n - 1 ? '\n' : ' ');
	system("pause");
	return 0;
}