交换二叉树的左右子树——非递归方式

来源：互联网发布：小鱼儿主页最近域名编辑：程序博客网时间：2024/06/14 01:02

这是华为的一道机试题，其实并不难，不让用递归可以用栈来解决，具体的代码如下：

#include <stdio.h>#include <stdlib.h>struct node{    char data ;    struct node* left ;    struct node* right ;};struct tree{    struct node* root ;};void tree_create( struct node** node){    char data;    scanf("%c" , &data);    if(data == '#' ){        *node = NULL;    }    else{        *node = ( struct node*)malloc(sizeof( struct node));        (*node)-> data = data;        tree_create(&(*node)-> left);        tree_create(&(*node)-> right);    }}void tree_destory( struct node** node){    if(*node != NULL){        tree_destory(&(*node)-> left);        tree_destory(&(*node)-> right);        free(node);    }}void tree_first( struct node* node){    if(node == NULL)        return;    printf("%c " , node->data);    tree_first(node-> left);    tree_first(node-> right);}static struct node* stack[5] = {0};static int pos = -1;void push( struct node* node){    stack[++pos] = node;}void pop(){    --pos;}int empty(){    return pos == -1;}struct node* top(){    return stack[pos];}void exchange( struct node* node){    struct node* tnode = node;    struct node* tmp = NULL;    if(node == NULL)        return;    while(1){        tmp = tnode-> left;        tnode-> left = tnode->right ;        tnode-> right = tmp;        if(tnode->right ){            push(tnode-> right);        }        if(tnode->left ){            tnode = tnode->left;        }        else{            if(!empty()){                tnode = top();                pop();            }            else{                break;            }        }    }}int main(){    struct tree tree;    tree_create(&tree. root);    printf("交换前的先序遍历：" );    tree_first(tree. root);    printf("\n" );    exchange(tree. root);    printf("交换后的先序遍历：" );    tree_first(tree. root);    printf("\n" );    tree_destory(&tree. root);    return 0;}

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本文作者：girlkoo

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