Leetcode: Binary Tree Upside Down

Binary Tree Upside Down

Similar Problems:

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.

For example:

```Given a binary tree {1,2,3,4,5},
1
/ \
2   3
/ \
4   5
```

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

```  4
/ \
5   2
/ \
3   1
```

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/binary-tree-upside-down
## Basic Ideas: In-order
##
## Complexity: Time O(n), Space O(h). h = height of the tree
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def upsideDownBinaryTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if root is None: return None
stack = []
p = root
while p:
stack.append(p)
p = p.left
newRoot = stack.pop(-1)
q = newRoot
while len(stack) != 0:
pre_parent = stack.pop(-1)
q.left = pre_parent.right
q.right = pre_parent
q = pre_parent
# configure the last node as a leaf
q.left, q.right = None, None
return newRoot
```

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