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。
1 #include <iostream> 2 using namespace std; 3 4 struct Node 5 { 6 char data; 7 int left; 8 int right; 9 }; 10 11 int buildTree(Node tree[]); 12 bool isSame(Node tree1[], int root1, Node tree2[], int root2); 13 14 int main() 15 { 16 Node tree1[10], tree2[10]; 17 int root1, root2; 18 root1 = buildTree(tree1); 19 root2 = buildTree(tree2); 20 if (isSame(tree1, root1, tree2, root2)) cout << "Yes"; 21 else cout << "No"; 22 return 0; 23 } 24 25 int buildTree(Node tree[]) 26 { 27 int num, i, root = -1; 28 char d, l, r; 29 cin >> num; 30 int judgeRoot[10] = {}; 31 for (i = 0; i < num; i++) 32 { 33 cin >> d >> l >> r; 34 tree[i].data = d; 35 if (l != '-') 36 { 37 tree[i].left = l - '0'; 38 judgeRoot[l - '0'] = 1; 39 } 40 else tree[i].left = -1; 41 if (r != '-') 42 { 43 tree[i].right = r - '0'; 44 judgeRoot[r - '0'] = 1; 45 } 46 else tree[i].right = -1; 47 } 48 for (i = 0; i < num; i++) 49 { 50 if (judgeRoot[i] == 0) root = i; 51 } 52 return root; 53 } 54 55 bool isSame(Node tree1[], int r1, Node tree2[], int r2) 56 { 57 if (r1 == -1 && r2 == -1) return true; 58 if (r1 == -1 && r2 > -1 || r1 > -1 && r2 == -1) return false; 59 if (tree1[r1].data != tree2[r2].data) return false; 60 return isSame(tree1, tree1[r1].left, tree2, tree2[r2].left) 61 && isSame(tree1, tree1[r1].right, tree2, tree2[r2].right) 62 || isSame(tree1, tree1[r1].left, tree2, tree2[r2].right) 63 && isSame(tree1, tree1[r1].right, tree2, tree2[r2].left); 64 }