nvert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9

to

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:

This problem was inspired by
this original tweet by
Max Howell:

  Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

非递归算法：

1、交换根节点的左右子节点

2、交换第二层每个节点的左右子节点

….

这个与二叉树层次遍历类似，代码如下：

TreeNode* invertTree2(TreeNode* root) {
		queue<TreeNode*> tree_queue;
		if (root == NULL)
			return root;
		tree_queue.push(root);
		while(tree_queue.size() > 0){
			TreeNode * pNode = tree_queue.front();
			tree_queue.pop();
			TreeNode * pLeft = pNode->left;
			pNode->left = pNode->right;
			pNode->right = pLeft;
			if (pNode->left)
				tree_queue.push(pNode->left);
			if (pNode->right)
				tree_queue.push(pNode->right);
		}
		return root;
	}

递归算法：

1、交换根节点的左右子树。

2、对左右子树分别执行递归反转 。

代码如下：

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