剑指offer之树的子结构

1 题目

输入两颗二叉树A和B，判断B是不是A的子结构（B树是A树的子结构）
比如：
2
树A 3 5 树B 5
1 4 2 3 2 3
很明显树B是树A的子结构

2 代码实现

#include <stdio.h>#define true 1
#define false 0typedef struct Node
{int value;struct Node* left;struct Node* right;
} Node;int has_sub_tree(Node *head1, Node *head2)
{int result = false;if (head1 != NULL && head2 != NULL){printf("head1->value is %d\n", head1->value);printf("head2->value is %d\n", head2->value);if (head1->value == head2->value){result = is_same(head1, head2); }if (!result){result = has_sub_tree(head1->left, head2);}if (!result){result = has_sub_tree(head1->right, head2);}}return result;
}int is_same(Node *head1, Node *head2)
{if (head2 == NULL){return true;}if (head1 == NULL){return false;}printf("is_same head1->value is %d\n", head1->value);printf("is_same head2->value is %d\n", head2->value);if (head1->value != head2->value){return false;}return is_same(head1->left, head2->left) && is_same(head1->right, head2->right);
}void printf_tree(Node *head)
{if (head != NULL){printf("val is: %d\n", head->value);printf_tree(head->left);printf_tree(head->right);}
}int main()
{/*              2*           3    5            5*         1  4  2  3        2   3*       */Node head1, node1, node2, node3, node4, node5, node6;Node head2, node7, node8;head1.value = 2;node1.value = 3;node2.value = 5;node3.value = 1;node4.value = 4;node5.value = 2;node6.value = 3;head1.left = &node1;head1.right = &node2;node1.left = &node3;node1.right = &node4;node2.left = &node5;node2.right = &node6;node3.left = NULL;node3.right = NULL;node4.left = NULL;node4.right = NULL;node5.left = NULL;node5.right = NULL;node6.left = NULL;node6.right = NULL;head2.value = 5;node7.value = 2;node8.value = 3;head2.left = &node7;head2.right = &node8;node7.left = NULL;node7.right = NULL;node8.left = NULL;node8.right = NULL;printf_tree(&head1);printf_tree(&head2);int result = has_sub_tree(&head1, &head2);printf("result is %d\n", result);return 0;
}

3 运行结果

val is: 2
val is: 3
val is: 1
val is: 4
val is: 5
val is: 2
val is: 3
val is: 5
val is: 2
val is: 3
head1->value is 2
head2->value is 5
head1->value is 3
head2->value is 5
head1->value is 1
head2->value is 5
head1->value is 4
head2->value is 5
head1->value is 5
head2->value is 5
is_same head1->value is 5
is_same head2->value is 5
is_same head1->value is 2
is_same head2->value is 2
is_same head1->value is 3
is_same head2->value is 3
result is 1

4 总结

一开始is_same写错了，实现如下

int is_same(Node *head1, Node *head2)
{if (head1 == NULL){return false;}if (head2 == NULL){return true;}printf("is_same head1->value is %d\n", head1->value);printf("is_same head2->value is %d\n", head2->value);if (head1->value != head2->value){return false;}return is_same(head1->left, head2->left) && is_same(head1->right, head2->right);
}

这样写导致的错误就是，比如
2
树A 3 5 树B 5
1 4 2 3 2 3
树B的5节点和树A的5节点进行匹配，然后树B的2节点和树A的2节点进行匹配，接下来，树A的left是NULL了，直接返回false,那么后面的 && is_same(head1->right, head2->right)
就不会再执行了，所以返回false,然而B数的右结构没有进行比较是直接false了，所以我们需要把