【算法】输入一颗二元查找树，将该树转换为它的镜像

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题目：输入一棵二元查找树，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。

要求：用递归和循环两种方法完成树的镜像转换。

举例：

        8                     8      /   \       转换       /   \     6     10     -->      10    6    /  \   / \            /  \   / \   5    7 9   11         11   9 7   5

#include <iostream>#include <cstdio>#include <cstdlib>#include <queue>#include <stack>using namespace std;typedef struct node{int key;struct node *pleft;struct node *pright; }Node;int CreateTreeByInsertData(Node **p,int k)//理解为什么用二级指针{if(*p==NULL)      {*p=(Node *)malloc(sizeof(Node));(*p)->key=k; (*p)->pleft=(*p)->pright=NULL;     return 1;}else if(k == (*p)->key)          return 0;else if(k < (*p)->key)           return CreateTreeByInsertData(&(*p)->pleft,k); elsereturn CreateTreeByInsertData(&(*p)->pright,k);  }void swap(Node **l, Node **r){Node *p = *l;*l = *r;*r = p;}void mirror(Node *root)//递归来使得镜像对称{if(root == NULL)return;swap( &(root->pleft), &(root->pright));mirror(root->pleft);mirror(root->pright);}void mirrorByStack(Node *root)//用栈来做得镜像对称{stack<Node *>myStack;if(root == NULL)return;myStack.push(root);while(!myStack.empty()){Node *p = myStack.top();//取得栈顶元素myStack.pop();//只是弹出，删除，并不能取到栈顶元素swap( &(p->pleft), &(p->pright));if(p->pleft!=NULL)myStack.push(p->pleft);if(p->pright!=NULL)myStack.push(p->pright);}}void visitByLevel(Node *p)//层次遍历，上到下，左到右{queue<Node*> myQueue;if(p == NULL)return;myQueue.push(p);while(!myQueue.empty()){Node *now = myQueue.front();myQueue.pop();printf("%d ",now->key);if(now->pleft) myQueue.push(now->pleft);if(now->pright) myQueue.push(now->pright);}printf("\n");}void ClearTree(Node** tree)//删除树的操作，在本题中不一定用的到{if(*tree==NULL)return;ClearTree(&(*tree)->pleft);ClearTree(&(*tree)->pright);free(*tree);*tree=NULL;}int main(){int i;Node *proot = NULL;int data[] = {8,6,10,5,7,9,11};//依次插入一些数据，创建一个二叉排序树for(i=0; i<sizeof(data)/sizeof(int); i++)CreateTreeByInsertData(&proot, data[i]);visitByLevel(proot);//8 6 10 5 7 9 11mirror(proot);visitByLevel(proot);//8 10 6 11 9 7 5mirrorByStack(proot);visitByLevel(proot);//8 6 10 5 7 9 11ClearTree(&proot);return 0; }