LeetCode之Increasing Order Search Tree（Kotlin）

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()
        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) {
        if (root == null) {
            return
        }
        inorder(root.left, list)
        list.add(root)
        inorder(root.right, list)
    }
}

有问题随时沟通

具体代码实现可以参考Github