Flatten Binary Tree to Linked List

LeetCode 114 | Difficulty: Medium

Medium

Problem 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**](https://en.wikipedia.org/wiki/Tree_traversal#Pre-order,_NLR) 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)?

Topics: Linked List, Stack, Tree, Depth-First Search, Binary Tree

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Approach

Tree DFS

Traverse the tree recursively (or with a stack). At each node, decide: what information do I need from the left/right subtrees? Process: go left → go right → combine results. Consider preorder, inorder, or postorder traversal based on when you need to process the node.

When to use

Path problems, subtree properties, tree structure manipulation.

Stack

Use a stack (LIFO) to track elements that need future processing. Process elements when a "trigger" condition is met (e.g., finding a smaller/larger element). Monotonic stack maintains elements in sorted order for next greater/smaller element problems.

When to use

Matching brackets, next greater element, evaluating expressions, backtracking history.

Linked List

Use pointer manipulation. Common techniques: dummy head node to simplify edge cases, fast/slow pointers for cycle detection and middle finding, prev/curr/next pattern for reversal.

When to use

In-place list manipulation, cycle detection, merging lists, finding the k-th node.

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Solutions

Solution 1: C# (Best: 168 ms)

Metric	Value
Runtime	168 ms
Memory	36.6 MB
Date	2022-01-13

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public void Flatten(TreeNode root) {
        TreeNode current = root;
        while(current != null)
        {
            if(current.left != null) {
            TreeNode prev = current.left;
            while(prev.right != null)
            {
                prev = prev.right;
            }
            prev.right = current.right;
            current.right = current.left;
            current.left = null;
            }
            current = current.right;
        }
        
    }
}

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Complexity Analysis

Approach	Time	Space
Tree Traversal	$O(n)$	$O(h)$
Stack	$O(n)$	$O(n)$
Linked List	$O(n)$	$O(1)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.
• Draw the pointer changes before coding. A dummy head node simplifies edge cases.
• Think about what triggers a pop: is it finding a match, or finding a smaller/larger element?
• LeetCode provides 1 hint(s) for this problem — try solving without them first.

💡 Hints

Hint 1: If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.