Problem

Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.

Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

5
/ \
3    6
/ \    \
2   4    8
/        / \
1        7   9

Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

1
\
2
\
3
\
4
\
5
\
6
\
7
\
8
\
9

Note:

1. The number of nodes in the given tree will be between 1 and 100.
2. Each node will have a unique integer value from 0 to 1000.

Solution: In-order traversal

root = 5
inorder(root.left) 之后
self.prev = 4
（1-4）已经处理完了，这时候的树是很奇怪的一个形状，3即是2的右子树，又是5的左子树。
1
\
2    5
\  /  \
3     6
\     \
prev -> 4     8
/  \
7    9
—————————
5.left = None # 把5->3的链接断开
5
\
6
\
8
/  \
7    9
—————————–
self.prev.right = root  <=> 4.right = 5

1
\
2
\
3
\
4
\
5 <– prev
\
6
\
8
/   \
7     9
self.prev = root <=> prev = 5
inorder(5.right) <=> inorder(6) 然后再去递归处理6（及其子树）即可。

Time complexity: O(n)

Space complexity: O(n)

Python3

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