数据结构之线索二叉树的简单实现

二叉树的链式存储时会有较多空间的浪费，当一颗有 n 个节点的二叉树存储共有2n个指针域，但是只有n-1个被用到，所以剩下的n+1个都会被浪费掉，因此可以想一种办法把剩余的空间利用起来，线索二叉树应运而生。
将二叉树重新定义如下每个节点为下列形式
lchild ltag data rtag rchild
其中：ltag=0 时lchild指向左子女；
ltag=1 时lchild指向前驱；
rtag=0 时rchild指向右子女；
rtag=1 时rchild指向后继；
所以结构体可定义为：

typedef enum{
	Link,//枚举，第一个默认为0，后面的自动 +1
	Thread
}PointerTag;

typedef struct Tree_Node{
	char ch;
	struct Tree_Node *left;
	struct Tree_Node *right;
	PointerTag ltag;
	PointerTag rtag;
	
}Tree;

中序线索化

//中序线索化并实现打印
void midThread(Tree *tree){
	if(tree){
		midThread(tree->left);
		printf("%c ",tree->ch);
		if(!tree->left){
			tree->ltag = Thread;
			tree->left = pre;
		}
		if(!tree->right)
		tree->rtag = Thread;
		if(pre && pre->rtag == Thread) {
			pre->right = tree;
		}
		pre = tree;
		midThread(tree->right);
	}
} 

后序线索化

void end(Tree *tree){
	if(tree){
		end(tree->left);
		end(tree->right);
		printf("%c ",tree->ch);
		if(!tree->left){
			tree->ltag = Thread;
			tree->left = pre;
		}
		if(!tree->right)
		tree->rtag = Thread;
		if(pre && pre->rtag == Thread)
		pre->right = tree;
		pre = tree;
	}
}

完整代码如下

#include <stdio.h>
#include <stdlib.h>

typedef enum{
	Link,
	Thread
}PointerTag;

typedef struct Tree_Node{
	struct Tree_Node *left;
	struct Tree_Node *right;
	char ch;
	PointerTag ltag;
	PointerTag rtag;
}Tree;

Tree *pre;

Tree *create(){
	Tree *tree;
	char ch;
	scanf("%c",&ch);
	if(ch == '*')
	tree = NULL;
	else{
		tree = (Tree *)malloc(sizeof(Tree));
		tree->ch = ch;
		tree->left = create();
		tree->right = create();
	}
	return tree;
}

void midThread(Tree *tree){
	if(tree){
		midThread(tree->left);
		printf("%c ",tree->ch);
		if(!tree->left){
			tree->ltag = Thread;
			tree->left = pre;
		}
		if(!tree->right)
		tree->rtag = Thread;
		if(pre && pre->rtag == Thread)
		pre->right = tree;
		pre = tree;
		midThread(tree->right);
	}
}



void end(Tree *tree){
	if(tree){
		end(tree->left);
		end(tree->right);
		printf("%c ",tree->ch);
		if(!tree->left){
			tree->ltag = Thread;
			tree->left = pre;
		}
		if(!tree->right)
		tree->rtag = Thread;
		if(pre && pre->rtag == Thread)
		pre->right = tree;
		pre = tree;
	}
}
int main(){
	Tree *tree = create();
	midThread(tree); 
	//end(tree);//调用时只能一个一个调用
}