7-3 homogeneous tree (25 minutes)
Given two trees T1 and T2. If T1 can be interchanged by several times about the child becomes T2, the two trees we call a "homogeneous" is. For example, Figure 1 shows two trees is isomorphic, as we have a tree in which the nodes A, B, G around the child interchange, get another tree. And FIG. 2 is not homogeneous.
figure 1
figure 2
Now given the two trees, you determine whether they are isomorphic.
Input formats:
2 input information given binary tree. For each tree, the first row is given a non-negative integer N (≦ 10), i.e., the number of nodes in the tree (assuming this case from node number 0 to N-1); followed by N rows, corresponding to the i-th row No. i-th node, the given node is stored in a capital letters, its left child node number, right child node numbers. If the child node is empty, then given the corresponding position "-." Separated by a space between the data given. Note: Title ensure that each node is stored in a different letter.
Output formats:
If two trees are isomorphic, output "Yes", otherwise a "No".
Input Sample 1 (corresponding to FIG. 1):
8
A 1 2
B 3 4
C 5 -
D - -
E 6 -
G 7 -
F - -
H - -
8
G - 4
B 7 6
F - -
A 5 1
H - -
C 0 -
D - -
E 2 -
Output Sample 1:
Yes
Input Sample 2 (corresponding to FIG. 2):
8
B 5 7
F - -
A 0 3
C 6 -
H - -
D - -
G 4 -
E 1 -
8
D 6 -
B 5 -
E - -
H - -
C 0 2
G - 3
F - -
A 1 4
Output Sample 2:
No
#include<iostream>
#include <vector>
#include<string>
#include <map>
using namespace std;
struct Node{
char* left,* right;
char l, r;
};
void input(map<char, Node>& tree) {
int n;
string a, b;
cin >> n;
vector<char> nodes(n);
for (int i = 0; i < n; ++i) {
cin >> nodes[i] >> a >> b;
tree[nodes[i]] ={ a == "-" ? nullptr : &nodes[stoi(a)] ,b == "-" ? nullptr : &nodes[stoi(b)] };
}
for (auto& it : tree) //不能用指针做为参考，指针出函数就无效
it.second.l = it.second.left!=nullptr?*it.second.left:'-', it.second.r = it.second.right!= nullptr ? *it.second.right: '-';
}
bool judge(map<char, Node>& a, map<char, Node>& b) {
if (a.size() != b.size())
return false;
for (auto& it : a) {
if (it.second.l != b[it.first].r && it.second.r != b[it.first].l&& //互换
(it.second.l != b[it.first].l && it.second.r != b[it.first].r)) //不互换
return false;
}
return true;
}
int main(){
map<char, Node> a, b;
input(a); input(b);
cout << (judge(a, b)?"Yes":"No");
return 0;
}