【PAT】1119. Pre- and Post-order Traversals (30)

1119. Pre- and Post-order Traversals (30)

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

分析：二叉树左右子树有序，当只有1个子树时，无法唯一确定（判断点）；

　　　两个子树初始位置指针+1个相应子树长度表示该子树的排序，递归分析子树；

细节：最后要输出“回车”（坑点）

#include <iostream>
#include <queue>
#include <stack>

using namespace std;

const int Max=31;
int temp=0;
queue<int> q;

struct node
{
	int info;
	int left;
	int right;
}tree[Max];

int pre[Max];
int post[Max];

int creat(int n)
{
	for (int i=0;i<Max;++i){
		tree[i].info=tree[i].left=tree[i].right=0;
	}
	for (int i=1;i<=n;++i) cin>>pre[i];
	for (int i=1;i<=n;++i) cin>>post[i]; 
	return post[n];
}

void build(int root, int p1, int p2, int n)
{
	tree[root].info=root;
	if (n==1) return;
	int l=pre[p1+1];
	int r=post[p2-1];
	if (l==r) {
		temp=1;
		tree[root].right=r;
		if (n>2) build(l,p1+1,p2-1,n-1);
		return;
	}
	tree[root].left=l;
	tree[root].right=r;
	for (int i=p1;i<=n+p1;++i){
		if (pre[i]==post[p2-1]){
			build (l,p1+1,p2-n+i-p1-1,i-p1-1);
			build(r,i,p2-1,n-i+p1);
			break;
		}
	}
}

void inorder(int root)
{
	if (tree[root].left) inorder(tree[root].left);
	q.push(root);
	if (tree[root].right) inorder(tree[root].right);
}

void print(int root)
{
	if (temp==0) cout<<"Yes\n";
	else cout<<"No\n";
	inorder(root);
	while (!q.empty()){
		cout<<q.front();
		q.pop();
		if (!q.empty()) cout<<' ';
	}
	cout<<endl;
}

int main()
{
//	freopen("test.txt","r",stdin);
	int N;
	cin>>N;
	int root;
	root=creat(N);
	build(root,1,N,N);
	print(root);
	return 0;
}