微软算法面试(10):输入一颗二元查找树,将该树转换为它的镜像
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。
例如:
8
/ /
6 10
// //
5 7 9 11
输出:
8
/ /
10 6
// //
11 9 7 5
分析:
可以使用递归思想,BTree* root ; 先转换左子树(root->left), 再转换右子树(root->right),最后把左子树的结果和右子树的结果对换下,就OK了。
很简单吧,看看实现:
#include<iostream>
using namespace std;
struct BTree{
BTree(int _v):value(_v), left(NULL), right(NULL) {}
int value;
BTree* left;
BTree* right;
void add(BTree* _bt)
{
if(_bt->value < value)
{
if(left == NULL)
left = _bt;
else
left->add(_bt);
}
else
{
if(right == NULL)
right = _bt;
else
right->add(_bt);
}
}
};
void transBTree(BTree* bt)
{
if(bt == NULL) return;
if(bt->left != NULL) transBTree(bt->left);
if(bt->right != NULL) transBTree(bt->right);
BTree* ptmp = bt->left;
bt->left = bt->right;
bt->right = ptmp;
}
template <typename T, int size=1024>
class queue{
private:
T data[size];
int begin, end;
public:
queue():begin(0),end(0){}
bool empty()
{
return begin == end;
}
void push(T& _d)
{
if((end +1)%size == begin)
{
cout << "queue is full, cann't push any node." <<endl;
return;
}
data[end] = _d;
end = (end + 1)%size;
}
T& pop(T& _t)
{
if(end == begin)
{
cout << "queue is empty, cann't pop any node." <<endl;
return _t;
}
_t = data[begin];
begin = (begin + 1)%size;
return _t;
}
};
void print(BTree* root)
{
queue<BTree*> q;
BTree* node = root;
BTree _n1(-1);//空标记
q.push(node);
while(!q.empty())
{
BTree* _bsn = &_n1;
q.pop(_bsn);
cout << _bsn->value << " ";
if(_bsn->left != NULL)
q.push(_bsn->left);
if(_bsn->right != NULL)
q.push(_bsn->right);
}
}
int main()
{
BTree* root;
BTree bt1(8);
BTree bt2(6);
BTree bt3(10);
BTree bt4(5);
BTree bt5(7);
BTree bt6(9);
BTree bt7(11);
root = &bt1;
root->add(&bt2);
root->add(&bt3);
root->add(&bt4);
root->add(&bt5);
root->add(&bt6);
root->add(&bt7);
cout << "按层次,从左到右打印结果为:";
print(root);
cout << endl;
transBTree(root);
cout << "转换后:按层次,从左到右打印结果为:";
print(root);
return 0;
}
输出结果为:
按层次,从左到右打印结果为:8 6 10 5 7 9 11
转换后:按层次,从左到右打印结果为:8 10 6 11 9 7 5