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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;
}
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原文地址：http://www.cnblogs.com/jasonkent27/p/4824844.html
