劍指Offer 1521 二叉樹的鏡像

第15題：
題目：輸入一顆二元查找樹，將該樹轉換爲它的鏡像，
即在轉換後的二元查找樹中，左子樹的結點都大於右子樹的結點。
用遞歸和循環兩種方法完成樹的鏡像轉換。   

#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;
}