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114. Flatten Binary Tree to Linked List

Medium

Given the root of a binary tree, flatten the tree into a “linked list”:

• The “linked list” should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
• The “linked list” should be in the same order as a pre-order traversal of the binary tree.

Example 1:

Input: root = [1,2,5,3,4,null,6]

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

Example 2:

Input: root = []

Output: []

Example 3:

Input: root = [0]

Output: [0]

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -100 <= Node.val <= 100

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

Solution

import com_github_leetcode.TreeNode

/*
 * Definition for a binary tree node.
 * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {
 *   var value: Int = _value
 *   var left: TreeNode = _left
 *   var right: TreeNode = _right
 * }
 */
object Solution {
    def flatten(root: TreeNode): Unit = {
        if (root != null) {
            modifyToRight(root)
        }
    }

    private def modifyToRight(root: TreeNode): TreeNode = {
        if (root.left == null && root.right == null) {
            root
        } else if (root.left == null) {
            root.right = modifyToRight(root.right)
            root
        } else if (root.right == null) {
            val left = modifyToRight(root.left)
            root.left = null
            root.right = left
            root
        } else {
            val left = modifyToRight(root.left)
            val right = modifyToRight(root.right)
            traverseAndAppend(left, right)
            root.right = left
            root.left = null
            root
        }
    }

    private def traverseAndAppend(root: TreeNode, right: TreeNode): Unit = {
        if (root.right != null) {
            traverseAndAppend(root.right, right)
        } else {
            root.right = right
        }
    }
}

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