0114. Flatten Binary Tree to Linked List

https://leetcode.com/problems/flatten-binary-tree-to-linked-list

Description

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)?

ac

class Solution {
    public void flatten(TreeNode root) {
        if (root == null || (root.left == null && root.right == null)) return;

        flatten(root.left);
        flatten(root.right);

        TreeNode tmpR = root.right;
        TreeNode tmpL = root.left;
        root.left = null;
        root.right = tmpL;
        while (tmpL.right != null)  tmpL = tmpL.right;
        tmpL.right = tmpR;

        return;
    }
}

ac2

so smart! you can start from leftmost, but you can start from rightmost as well!

class Solution {
    private TreeNode tmp = null;
    public void flatten(TreeNode root) {
        if (root == null) return;

        flatten(root.right);
        flatten(root.left);

        root.left = null;
        root.right = tmp;
        tmp = root;
    }
}