问题：
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:
The number of nodes in the given tree will be between 1 and 100.
Each node will have a unique integer value from 0 to 1000.
方法：
因为是二叉搜索树，所以中序遍历之后得到的序列就是从小到大的节点排序。然后从新生成树结构即可得答案。
具体实现：
class IncreasingOrderSearchTree {
// Definition for a binary tree node.
class TreeNode(var `val`: Int = 0) {
var left: TreeNode? = null
var right: TreeNode? = null
}
fun increasingBST(root: TreeNode?): TreeNode? {
val list = mutableListOf<TreeNode>()
inorder(root, list)
for (index in 0 until list.lastIndex) {
list[index].left = null
list[index].right = list[index + 1]
}
list[list.lastIndex].left = null
list[list.lastIndex].right = null
val result = list[0]
list.clear()
return result
}
private fun inorder(root: TreeNode?, list: MutableList<TreeNode>) {
if (root == null) {
return
}
inorder(root.left, list)
list.add(root)
inorder(root.right, list)
}
}
有问题随时沟通
