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:

71 2 3 4 6 7 52 6 7 4 5 3 1

Sample Output 1:

Yes2 1 6 4 7 3 5

Sample Input 2:

41 2 3 42 4 3 1

Sample Output 2:

No2 1 3 4

由于对树的遍历序列缺乏深入了解，所以用了费力不讨好的穷举方式，模仿前（后）序和中序序列建树的方式，对剩下部分进行集合划分，来穷举所有情况。但为了保留一个解，所以找到一个解和判断解是否唯一用了两次建树（暴露智商）,因为不知道在递归建树的过程中能不能确定一个树是否枚举完成（如果有的话请赐教），如果能的话就不用第二个函数去计算有多少种树。而且不知道给定两个序列最多有多少种树，所以也不能保证是否溢出结果是否正确。但实际运行的结果是有两个测试点段错误，不知道是不是数组越界还是栈溢出。

简便方法参考了其他人的方法。

自己的：

#include <iostream>#include <vector>#include <unordered_set>#include <map>using namespace std;struct node{int left,right,data;node(int d = 0,int l=-1,int r=-1):left(l),right(r){}};int id = 0,kase = 0,n;node tree[35];int pre[35],post[35];vector<int> ans;bool same(const unordered_set<int> & v1,const unordered_set<int> & v2){for(auto itr = v1.begin();itr != v1.end();++itr){if(v2.find(*itr) == v2.end())return false;}return true;}void inorder(int root){if(root != -1){inorder(tree[root].left);ans.push_back(tree[root].data);inorder(tree[root].right);}}int build_tree(int a1,int b1,int a2,int b2,bool & f,int level){int r= -1;f = false;if(b1 - a1 != 0){if(pre[a1] != post[b2-1]){f = false;return -1;}unordered_set<int> v1,v2;int r = id++,flag=0;tree[r].data = pre[a1];bool f1 = false,f2= false;tree[r].left = -1;f2=true;tree[r].right = build_tree(a1+1,b1,a2,b2-1,f2,level+1);if(f1&&f2){f = true;if(level == 0){inorder(0);}return r;}for(int i=0;i<b1-a1-1;i++){v1.insert(pre[a1+i+1]);v2.insert(post[a2+i]);if(same(v1,v2)==true){tree[r].left = build_tree(a1+1,a1+1+i+1,a2,a2+i+1,f1,level+1);tree[r].right = build_tree(a1+1+i+1,b1,a2+i+1,b2-1,f2,level+1);if(f1&&f2){f = true;if(level == 0){inorder(0);}return r;}}}tree[r].right = -1;f2=true;tree[r].left = build_tree(a1+1,b1,a2,b2-1,f1,level+1);if(f1&&f2){f = true;if(level == 0){inorder(0);}return r;}}else {f = true;}return r;}int just_one(int a1,int b1,int a2,int b2,bool & f,int level,int & cnt){int r= -1;f = false;int c1=0,c2=0;cnt = 0;if(b1 - a1 != 0){if(pre[a1] != post[b2-1]){f = false;return -1;}unordered_set<int> v1,v2;int r = id++,flag=0;tree[r].data = pre[a1];bool f1 = false,f2= false;tree[r].left = -1;f2=true; c1= 1;tree[r].right = just_one(a1+1,b1,a2,b2-1,f2,level+1,c2);if(f1&&f2){f = true;cnt += c1*c2;}for(int i=0;i<b1-a1-1;i++){v1.insert(pre[a1+i+1]);v2.insert(post[a2+i]);if(same(v1,v2)==true){tree[r].left = just_one(a1+1,a1+1+i+1,a2,a2+i+1,f1,level+1,c1);tree[r].right = just_one(a1+1+i+1,b1,a2+i+1,b2-1,f2,level+1,c2);if(f1&&f2){f = true;cnt += c1*c2;}}}tree[r].right = -1;f2=true;c2 = 1;tree[r].left = just_one(a1+1,b1,a2,b2-1,f1,level+1,c1);if(f1&&f2){f = true;cnt += c1*c2;}}else {f = true;cnt = 1;}return r;}void display(const vector<int> & vec){for(int i=0;i<vec.size();i++){if(i != 0)cout<<" ";cout<<vec[i];}cout<<endl;}int main(){cin>>n;for(int i=0;i<n;i++){cin>>pre[i];}for(int i=0;i<n;i++){cin>>post[i];}bool f;int r = build_tree(0,n,0,n,f,0);int cnt = 0;just_one(0,n,0,n,f,0,cnt);if(cnt == 1)cout<<"Yes\n";else cout<<"No\n";display(ans);return 0;}

参考了http://blog.csdn.net/liuchuo/article/details/52505179

#include <iostream>#include <vector>using namespace std;bool one = true;vector<int> res;int pre[35],post[35],n;//已经确定是合法的前后序遍历序列，所以不用判断是否合法。void handle(int a1,int b1,int a2,int b2){if(b1-a1>1){int idx = 0;for(int i=0;i<b1-a1-1;i++)if(pre[a1+1+i] == post[b2-2]){idx = i;break;}if(idx > 0){handle(a1+1,a1+1+idx,a2,a2+idx);res.push_back(pre[a1]);handle(a1+1+idx,b1,a2+idx,b2-1);}else{one = false;res.push_back(pre[a1]);handle(a1+1,b1,a2,b2-1);}}else if(b1-a1 == 1 )res.push_back(pre[a1]);}int main(){cin>>n;for(int i=0;i<n;i++)cin>>pre[i];for(int i=0;i<n;i++)cin>>post[i];handle(0,n,0,n);if(one == true)cout<<"Yes\n";else cout<<"No\n";for(int i=0;i<res.size();i++){if(i != 0)cout<<" ";cout<<res[i];}cout<<endl;return 0;}