剑指Offer 1521 二叉树的镜像

来源：互联网发布：路由器mac地址编辑：程序博客网时间：2024/05/18 00:19

第15题：
题目：输入一颗二元查找树，将该树转换为它的镜像，
即在转换后的二元查找树中，左子树的结点都大于右子树的结点。
用递归和循环两种方法完成树的镜像转换。

#include <cstdlib>#include <iostream>#include <stdio.h>using namespace std;struct TreeNode{       TreeNode *left;       TreeNode *right;       int value;};bool firstprint;void change(TreeNode *current){     if(current->left == NULL && current->right == NULL)         return;     if(current->left != NULL)         change(current->left);     if(current->right != NULL)         change(current->right);     if(current->left != NULL || current->right != NULL)     {         TreeNode *temp = current->left;         current->left = current->right;         current->right = temp;     }}/*void change(TreeNode *root){    if(root==NULL)    {        return;    }    TreeNode *temp = root->left;    root->left = root->right;    root->right = temp;    if(root->left)        change(root->left);    if(root->right)        change(root->right);}*/void preTravel(TreeNode *current){     if(current){        if(firstprint == true) {            printf("%d",current->value);            firstprint = false ;        }        else printf(" %d",current->value);        preTravel(current->left);        preTravel(current->right);    }}int main(int argc, char *argv[]){    TreeNode node[1500];    int n;    int one, two;    char ci;    while(scanf("%d", &n) != EOF)    {        firstprint = true;        for(int i = 0; i < n; i++)            cin >> node[i].value;        for(int i = 0; i < n; i++)        {            cin >> ci;            if(ci == 'd')            {                cin >> one >> two;                node[i].left = &node[one-1];                node[i].right = &node[two-1];            }            else if(ci == 'l')            {                cin >> one;                node[i].left = &node[one-1];                node[i].right = NULL;            }            else if(ci == 'r')            {                cin >> two;                node[i].left = NULL;                node[i].right = &node[two-1];            }            else            {                node[i].left = NULL;                node[i].right = NULL;            }        }        if(n == 0)            printf("NULL\n");        else        {            change(&node[0]);            preTravel(&node[0]);            printf("\n");        }    }    return 0;}