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;
}
};

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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