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Description
Invert 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.
这道题让我们翻转二叉树，是树的基本操作之一，不算难题。最下面那句话实在有些木有节操啊，不知道是Google说给谁的。反正这道题确实难度不大，可以用递归和非递归两种方法来解。先来看递归的方法，写法非常简洁，五行代码搞定，交换当前左右节点，并直接调用递归即可，代码如下：
// Recursion
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if (!root) return NULL;
TreeNode *tmp = root->left;
root->left = invertTree(root->right);
root->right = invertTree(tmp);
return root;
}
};
非递归的方法也不复杂，跟二叉树的层序遍历一样，需要用queue来辅助，先把根节点排入队列中，然后从队中取出来，交换其左右节点，如果存在则分别将左右节点在排入队列中，以此类推直到队列中木有节点了停止循环，返回root即可。代码如下：
// Non-Recursion
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if (!root) return NULL;
queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
TreeNode *node = q.front(); q.pop();
TreeNode *tmp = node->left;
node->left = node->right;
node->right = tmp;
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
return root;
}
};
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