二叉树的拷贝、释放

快来打我* 2022-03-30 08:25 164阅读 0赞

![在这里插入图片描述][20190105141001803.png]

## 代码 ##

// ConsoleApplication1.cpp: 定义控制台应用程序的入口点。
    //
    
    #include "stdafx.h"
    #include <stdio.h>
    #include <string.h>
    #include <stdlib.h>
    
    
    
    typedef struct BINARYNODE
    { 
    	char ch;
    	struct BINARYNODE* left;
    	struct BINARYNODE* right;
    
    }BinaryNode;
    
    
    
    
    void Recursion(BinaryNode* root) //递归遍历
    { 
    	if (root == NULL) return;
    
    	Recursion(root->left);
    	Recursion(root->right);
    	printf("%c", root->ch);
    
    }
    
    
    BinaryNode* CopyTree(BinaryNode* root) //拷贝
    { 
    	if (root == NULL)
    		return NULL;
    
    	root->left=CopyTree(root->left);
    	root->right=CopyTree(root->right);
    
    	BinaryNode* Node = (BinaryNode*)malloc(sizeof(BinaryNode));
    	Node->ch = root->ch;
    	Node->left = root->left;
    	Node->right = root->right;
    
    	return Node;
    }
    
    
    void FreeTree(BinaryNode* root)//释放
    { 
    	if (root == NULL)
    		return;
    	FreeTree(root->left);
    	FreeTree(root->right);
    
    	free(root);
    }
    
    int main(void)
    { 
    	BinaryNode b1, b2, b3, b4, b5, b6, b7, b8;
    
    	b1.ch = 'A'; b1.left = &b2; b1.right = &b3;
    	b2.ch = 'B'; b2.left = NULL; b2.right = &b4;
    	b3.ch = 'F'; b3.left = NULL; b3.right = &b5;
    	b4.ch = 'C'; b4.left = &b6; b4.right = &b7;
    	b5.ch = 'G'; b5.left = &b8; b5.right = NULL;
    	b6.ch = 'D'; b6.left = NULL; b6.right = NULL;
    	b7.ch = 'E'; b7.left = NULL; b7.right = NULL;
    	b8.ch = 'H'; b8.left = NULL; b8.right = NULL;
    
    	BinaryNode* root = CopyTree(&b1); //将上面创建的二叉树拷贝给root（根）
    	Recursion(root);
    	FreeTree(root);
    
    	return 0;
    }

结果：  
![在这里插入图片描述][20190105141507259.png]

[20190105141001803.png]: /images/20220330/39118757e84e4e49823af5589e29fd2c.png
[20190105141507259.png]: /images/20220330/1198dbf922e0445385457d321634ccef.png