226. Invert Binary Tree

1. 问题描述

Invert a binary tree.

        4
       / 
      2   7
     /   / 
    1   3 6  9

to

      4
     / 
    7  2
   /   / 
  9   6 3  1                

Tag: Tree

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

2. 解题思路

3. 代码

class Solution 
{
public:
    TreeNode* invertTree(TreeNode* root)//递归实现
    {
        if (NULL == root)
        {
            return root;
        }
        if (NULL != root->left && NULL != root->right)
        {
            TreeNode *pt = root->right;
            root->right = root->left;
            root->left = pt;
            invertTree(root->left);
            invertTree(root->right);
            return root;
        }
    }
    TreeNode* invertTree2(TreeNode* root)//递归实现
    {
        if (NULL == root)
        {
            return root;
        } 
        TreeNode *pt = root->left;  
        root->left = invertTree2(root->right);  
        root->right = invertTree2(pt);  
        return root; 
    }
    TreeNode* invertTree3(TreeNode* root)//非递归方式：利用栈实现
    {
        if(NULL == root) 
        {
            return NULL;
        }
        std::stack<TreeNode *> tSta;
        tSta.push(root);
        while (!tSta.empty())
        {            
            TreeNode *pCur = tSta.top();
            tSta.pop();
            TreeNode *pt = pCur->left;
            pCur->left = pCur->right;
            pCur->right = pt;
            if (NULL != pCur->left)
            {
                tSta.push(pCur->left);
            }
            if (NULL != pCur->right)
            {
                tSta.push(pCur->right);
            }            
        }
        return root;  
    }
};

4. 反思