C++实现LeetCode(156.二叉树的上下颠倒)

[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：

									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：

									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;

									    }

									};