二叉树镜像

【题目描述】

操作给定的二叉树，将其变换为源二叉树的镜像。 

输入描述:

二叉树的镜像定义：源二叉树 
    	    8
    	   /  
    	  6   10
    	 /   / 
    	5  7 9 11
    	镜像二叉树
    	    8
    	   /  
    	  10   6
    	 /   / 
    	11 9 7  5

 

【代码实现】

方法1：递归实现

/*
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|| (pRoot->left==NULL && pRoot->right==NULL))
            return;
        TreeNode * node=pRoot->left;
        pRoot->left=pRoot->right;
        pRoot->right=node;
        if(pRoot->left)
            Mirror(pRoot->left);
        if(pRoot->right)
            Mirror(pRoot->right);
    }
};

方法二：迭代实现

/*
struct TreeNode {
    int val;//数据
    struct TreeNode *left;//左子节点指针
    struct TreeNode *right;//右子节点指针
    TreeNode(int x) ://用初始化列表进行初始的构造函数
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
public:
    void Mirror(TreeNode *pRoot) {
    if(pRoot==NULL|| (pRoot->left==NULL && pRoot->right==NULL))
            return;
        stack<TreeNode*> stackNode;
        stackNode.push(pRoot);
        while(stackNode.size()!=0){
            TreeNode* treeNode=stackNode.top();
            stackNode.pop();
            if(treeNode->left!=NULL ||treeNode->right!=NULL){
                TreeNode *temp=treeNode->left;
               treeNode->left=treeNode->right;
                treeNode->right=temp;
            }
            if(treeNode->left)
                stackNode.push(treeNode->left);
            if(treeNode->right)
                stackNode.push(treeNode->right);
        }
    }
};