114. Flatten Binary Tree to Linked List

1. 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.

2. Example

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]

3. Constraints

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

4. Solutions

Look for the Previous Node

n is the number of nodes in the tree
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    void flatten(TreeNode *root) {
        TreeNode *iter = root;
        while (iter != nullptr) {
            if (iter->left != nullptr && iter->right != nullptr) {
                auto prev = iter->left;
                while (prev->right != nullptr) {
                    prev = prev->right;
                }

                prev->right = iter->right;
            }

            if (iter->left != nullptr) {
                iter->right = iter->left;
                iter->left = nullptr;
            }

            iter = iter->right;
        }
    }
};