pta 编程题6 树的同构

其它pta数据结构编程题请参见：pta

题目请参见：树的同构

因题目中左右子树是按照下标给出，因此用数组存放树是更好的方法。

判断两棵树是否同构：用递归的方法。如果当前两个结点都为空，则返回TRUE；如果一个结点空一个不空，或者两个结点都不为空但是结点存储的数据不同，则返回FALSE。然后递归地的调用函数，判断是否左子树和左子树相同且右子树和右子树相同，或者左子树和左子树相同且右子树和右子树相同。

 #include <iostream>
 using namespace std;

 struct Node
 {
     char data;
     int left;
     int right;
 };

 int buildTree(Node tree[]);
 bool isSame(Node tree1[], int root1, Node tree2[], int root2);

 int main()
 {
     Node tree1[], tree2[];
     int root1, root2;
     root1 = buildTree(tree1);
     root2 = buildTree(tree2);
     if (isSame(tree1, root1, tree2, root2)) cout << "Yes";
     else cout << "No";
     return ;
 }

 int buildTree(Node tree[])
 {
     int num, i, root = -;
     char d, l, r;
     cin >> num;
     int judgeRoot[] = {};
     for (i = ; i < num; i++)
     {
         cin >> d >> l >> r;
         tree[i].data = d;
         if (l != '-')
         {
             tree[i].left = l - '';
             judgeRoot[l - ''] = ;
         }
         else tree[i].left = -;
         if (r != '-')
         {
             tree[i].right = r - '';
             judgeRoot[r - ''] = ;
         }
         else tree[i].right = -;
     }
     for (i = ; i < num; i++)
     {
         if (judgeRoot[i] == ) root = i;
     }
     return root;
 }

 bool isSame(Node tree1[], int r1, Node tree2[], int r2)
 {
     if (r1 == - && r2 == -) return true;
     if (r1 == - && r2 > - || r1 > - && r2 == -) return false;
     if (tree1[r1].data != tree2[r2].data) return false;
     return isSame(tree1, tree1[r1].left, tree2, tree2[r2].left)
         && isSame(tree1, tree1[r1].right, tree2, tree2[r2].right)
         || isSame(tree1, tree1[r1].left, tree2, tree2[r2].right)
         && isSame(tree1, tree1[r1].right, tree2, tree2[r2].left);
 }