# 156. Binary Tree Upside Down

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]`.

``````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;
}
};
``````

``````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

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

LeetCode All in One 题目讲解汇总(持续更新中…)

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