您的位置:首页 > 其它

[LeetCode] Invert Binary Tree 翻转二叉树

2015-06-13 00:54 489 查看
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;
}
};


LeetCode All in One 题目讲解汇总(持续更新中...)
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: