二叉树反转

invert a binary tree.

     4
   /   
  2     7
 /    / 
1   3 6   9

to

     4
   /   
  7     2
 /    / 
9   6 3   1


递归代码1:

TreeNode* invertTree(TreeNode* root) {
	if(root==NULL)
			return ;
		if ( root->left != NULL) invertTree(root->left);
                if ( root->right != NULL) invertTree(root->right);
		TreeNode * ptmpNode = root->left;
                root->left = root-right ;
                root->right = ptmpNode ;
    }        

  

 递归代码2:

   TreeNode* invertTree(TreeNode* root) {
	if(root==NULL)
			return NULL;
		TreeNode * ptmpNode = root->left;
		root->left = invertTree(root->right);
		root->right = invertTree(ptmpNode);
		return root;
    }

  

原文地址:https://www.cnblogs.com/jasonkent27/p/4824844.html