JSMS29-二叉树的镜像-剑指offer19

镜像二叉树，输入一个二叉树，输出二叉树的镜像。

分析：这道题不是很难,通过递归就可以实现。具体看代码

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

typedef struct d{
	int data;
	struct d *left;
	struct d *right;
}tree;

tree *buildt(int a[],int &n)
{
	int num=a[n++];
	if(num==NULL)
	{
		return NULL;
	}
	tree *s=(tree *)malloc(sizeof(tree));
	s->data=num;
	s->left=buildt(a,n);
	s->right=buildt(a,n);
	return s;
}

void Mirror(tree *t)             //主要函数
{
	if(t!=NULL)              //当tree！=NULL时候，
	{
		tree *tl,*tr;
		tl=t->left;      
		tr=t->right;     
		t->left=tr;     //左叶子指向t->right
		t->right=tl;    //右叶子指向t->left
		Mirror(t->left);    //递归左子树
		Mirror(t->right);   //递归右子树
	}else
	{
		return;
	}
}

tree *Ms(tree *t)                //这里Ms套个马甲，用于return 根节点
{
	tree *ret=t;
	Mirror(t);
	return ret;
}


void Pre(tree *t)
{
	if(t!=NULL)
	{
		printf("%d->",t->data);
		Pre(t->left);
		Pre(t->right);
	}
}

void Mid(tree *t)
{
	if(t!=NULL)
	{
		Mid(t->left);
		printf("%d->",t->data);
		Mid(t->right);
	}
}

int main()
{
	int a[]={1,2,4,0,0,5,0,0,3,6,0,0,7,0,0};
	int n=0;
	tree *t=buildt(a,n);
	printf("镜像前前序遍历:"); 
	Pre(t);
	printf("\n镜像前中序遍历");
	Mid(t);
	tree *t2=Ms(t);
	printf("\n镜像后前序遍历:");
	Pre(t2);
	printf("\n镜像后中序遍历:"); 
	Mid(t2); 
}