輸入描述:

二叉樹(shù)的鏡像定義：源二叉樹(shù) 
    	    8
    	   /      	  6   10
    	 / \  /     	5  7 9 11
    	鏡像二叉樹(shù)
    	    8
    	   /      	  10   6
    	 / \  /     	11 9 7  5

方法一

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    void Mirror(TreeNode *pRoot) {
        if(pRoot==NULL){
            return;
        }
        TreeNode *tmp = pRoot->left;
        pRoot->left = pRoot->right;
        pRoot->right = tmp;
        Mirror(pRoot->left);
        Mirror(pRoot->right);
    }
};

方法二：
class Solution {
public:
         //棧的非遞歸
     void Mirror(TreeNode *pRoot) {
         if (pRoot == NULL)return;
         stack<TreeNode*> st;
         TreeNode* p = NULL;
         st.push(pRoot);
         while (st.size())
         {
             p = st.top();
             st.pop();
             swap(p->left, p->right);
             if (p->left)st.push(p->left);
             if (p->right)st.push(p->right);
         }
     }  //隊(duì)列的非遞歸
    void Mirror(TreeNode *pRoot) {
         if (pRoot == NULL)return;
         queue<TreeNode*> qu;
         TreeNode* p = NULL;
         qu.push(pRoot);
         while (qu.size())
         {
             p = qu.front();
             qu.pop();
             swap(p->left, p->right);
             if (p->left)qu.push(p->left);
             if (p->right)qu.push(p->right);
         }
    }
    //遞歸
    void Mirror(TreeNode *pRoot) {
         if (pRoot == NULL)return;
         swap(pRoot->left, pRoot->right);
         Mirror(pRoot->left);
         Mirror(pRoot->right);
    }
};