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; } };
本文转自博客园Grandyang的博客,原文链接:翻转二叉树[LeetCode] Invert Binary Tree ,如需转载请自行联系原博主。