翻转二叉树 leetcode（二叉树 leetcode）

[LeetCode] 156. Binary Tree Upside Down 二叉树的上下颠倒

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

Example:

Input: [1,2,3,4,5]

1
/ \
2 3
/ \
4 5

Output: return the root of the binary tree [4,5,2,#,#,3,1]

4
/ \
5 2
/ \
3 1

Clarification:

Confused what[4,5,2,#,#,3,1]means? Read more below on how binary tree is serialized on OJ.

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

1
/ \
2 3
/
4
\
5

The above binary tree is serialized as[1,2,3,#,#,4,#,#,5].

这道题让我们把一棵二叉树上下颠倒一下，而且限制了右节点要么为空要么一定会有对应的左节点。上下颠倒后原来二叉树的最左子节点变成了根节点，其对应的右节点变成了其左子节点，其父节点变成了其右子节点，相当于顺时针旋转了一下。对于一般树的题都会有迭代和递归两种解法，这道题也不例外，先来看看递归的解法。对于一个根节点来说，目标是将其左子节点变为根节点，右子节点变为左子节点，原根节点变为右子节点，首先判断这个根节点是否存在，且其有没有左子节点，如果不满足这两个条件的话，直接返回即可，不需要翻转操作。那么不停的对左子节点调用递归函数，直到到达最左子节点开始翻转，翻转好最左子节点后，开始回到上一个左子节点继续翻转即可，直至翻转完整棵树，参见代码如下：

解法一：

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class Solution { public: TreeNode *upsideDownBinaryTree(TreeNode *root) { if (!root || !root->left) return root; TreeNode *l = root->left, *r = root->right; TreeNode *res = upsideDownBinaryTree(l); l->left = r; l->right = root; root->left = NULL; root->right = NULL; return res; } };

下面我们来看迭代的方法，和递归方法相反的时，这个是从上往下开始翻转，直至翻转到最左子节点，参见代码如下：

解法二：

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class Solution { public: TreeNode *upsideDownBinaryTree(TreeNode *root) { TreeNode *cur = root, *pre = NULL, *next = NULL, *tmp = NULL; while (cur) { next = cur->left; cur->left = tmp; tmp = cur->right; cur->right = pre; pre = cur; cur = next; } return pre; } };

Github 同步地址：

https://github.com/grandyang/leetcode/issues/156

类似题目：

Reverse Linked List

参考资料：

https://leetcode.com/problems/binary-tree-upside-down/

https://leetcode.com/problems/binary-tree-upside-down/discuss/49412/Clean-Java-solution

https://leetcode.com/problems/binary-tree-upside-down/discuss/49432/Easy-O(n)-iteration-solution-Java

https://leetcode.com/problems/binary-tree-upside-down/discuss/49406/Java-recursive-(O(logn)-space)-and-iterative-solutions-(O(1)-space)-with-explanation-and-figure

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原文链接：https://www.cnblogs.com/grandyang/p/5172838.html