剑指offer之面试题18树的子结构-优快云博客

本文链接：https://blog.youkuaiyun.com/WSYW126/article/details/51372186

本文提供了一种方法来判断一棵二叉树是否为另一棵二叉树的子结构，并附带了完整的C语言实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

问题描述：

输入两颗二叉树A和B，判断B是不是A的子结构。

实现代码如下：

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
struct BinaryTreeNode{
	int date;
	struct BinaryTreeNode * left;
	struct BinaryTreeNode * right;
};
bool doesTreeFhasTreeSon(struct BinaryTreeNode *TreeF,struct BinaryTreeNode *TreeSon){
	if(TreeSon==NULL)return true;
	if(TreeF==NULL)return false;
	
	if(TreeF->date!=TreeSon->date)return false;
	
	return doesTreeFhasTreeSon(TreeF->left,TreeSon->left) && doesTreeFhasTreeSon(TreeF->right,TreeSon->right);
	
}

bool hasSubTree(struct BinaryTreeNode *TreeF,struct BinaryTreeNode *TreeSon){
	bool result = false;
	if(TreeF !=NULL && TreeSon !=NULL){
		if(TreeF->date == TreeSon->date){
			result = doesTreeFhasTreeSon(TreeF,TreeSon);
		}
		if(!result){
			result = hasSubTree(TreeF->left,TreeSon);
		}
		if(!result){
			result = hasSubTree(TreeF->right,TreeSon);
		}
	}
	return result;	
}

int main(int argc, char *argv[])
{
	struct BinaryTreeNode *Fhead=(struct BinaryTreeNode *)malloc(sizeof(struct BinaryTreeNode));
	struct BinaryTreeNode *Sonhead=(struct BinaryTreeNode *)malloc(sizeof(struct BinaryTreeNode));
	int i;
	Fhead->date=0;
	Fhead->left=(struct BinaryTreeNode *)malloc(sizeof(struct BinaryTreeNode));
	Fhead->right=(struct BinaryTreeNode *)malloc(sizeof(struct BinaryTreeNode));
	Fhead->left->date=1;
	Fhead->left->left=NULL;
	Fhead->left->right=NULL;
	Fhead->right->date=2;
	Fhead->right->left=NULL;
	Fhead->right->right=NULL;
	Sonhead->date=2;
	Sonhead->left=NULL;
	Sonhead->right=NULL;
	
	bool b = hasSubTree(Fhead,Sonhead);
	b==true?printf("true"):printf("false");
	return 0;
}

参考资料：
剑指offer

备注：
转载请注明出处：http://blog.youkuaiyun.com/wsyw126/article/details/51372186
作者：WSYW126