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。
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。
例如输入：
8
/ \
6 10
/\ /\
5 7 9 11
输出：
8
/ \
10 6
/\ /\
11 9 7 5
思路：
左右子树交换
如果用循环来做：就是用辅助栈来模拟递归
// Mirror a BST (swap the left right child of each node) recursively// the head of BST in initial call///////////////////////////////////////////////////////////////////////void MirrorRecursively(BSTreeNode *pNode){ if(!pNode) return; // swap the right and left child sub-tree BSTreeNode *pTemp = pNode->m_pLeft; pNode->m_pLeft = pNode->m_pRight; pNode->m_pRight = pTemp; // mirror left child sub-tree if not null if(pNode->m_pLeft) MirrorRecursively(pNode->m_pLeft); // mirror right child sub-tree if not null if(pNode->m_pRight) MirrorRecursively(pNode->m_pRight); }程序基本上一样///////////////////////////////////////////////////////////////////////// Mirror a BST (swap the left right child of each node) Iteratively// Input: pTreeHead: the head of BST///////////////////////////////////////////////////////////////////////void MirrorIteratively(BSTreeNode *pTreeHead){ if(!pTreeHead) return; std::stack<BSTreeNode *> stackTreeNode; stackTreeNode.push(pTreeHead); while(stackTreeNode.size()) { BSTreeNode *pNode = stackTreeNode.top(); stackTreeNode.pop(); // swap the right and left child sub-tree BSTreeNode *pTemp = pNode->m_pLeft; pNode->m_pLeft = pNode->m_pRight; pNode->m_pRight = pTemp; // push left child sub-tree into stack if not null if(pNode->m_pLeft) stackTreeNode.push(pNode->m_pLeft); // push right child sub-tree into stack if not null if(pNode->m_pRight) stackTreeNode.push(pNode->m_pRight); }}