LeetCode #114 — MEDIUM

Flatten Binary Tree to Linked List

Move from brute-force thinking to an efficient approach using linked list strategy.

The Problem

Problem Statement

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)?

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Linked List · Stack · Tree

Example 1

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

Example 2

[]

Example 3

[0]

Core Insight

What unlocks the optimal approach

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

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #114: Flatten Binary Tree to Linked List
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public void flatten(TreeNode root) {
        while (root != null) {
            if (root.left != null) {
                // 找到当前节点左子树的最右节点
                TreeNode pre = root.left;
                while (pre.right != null) {
                    pre = pre.right;
                }

                // 将左子树的最右节点指向原来的右子树
                pre.right = root.right;

                // 将当前节点指向左子树
                root.right = root.left;
                root.left = null;
            }
            root = root.right;
        }
    }
}

// Accepted solution for LeetCode #114: Flatten Binary Tree to Linked List
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func flatten(root *TreeNode) {
	for root != nil {
		if root.Left != nil {
			pre := root.Left
			for pre.Right != nil {
				pre = pre.Right
			}
			pre.Right = root.Right
			root.Right = root.Left
			root.Left = nil
		}
		root = root.Right
	}
}

# Accepted solution for LeetCode #114: Flatten Binary Tree to Linked List
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        while root:
            if root.left:
                pre = root.left
                while pre.right:
                    pre = pre.right
                pre.right = root.right
                root.right = root.left
                root.left = None
            root = root.right

// Accepted solution for LeetCode #114: Flatten Binary Tree to Linked List
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    #[allow(dead_code)]
    pub fn flatten(root: &mut Option<Rc<RefCell<TreeNode>>>) {
        if root.is_none() {
            return;
        }
        let mut v: Vec<Option<Rc<RefCell<TreeNode>>>> = Vec::new();
        // Initialize the vector
        Self::pre_order_traverse(&mut v, root);
        // Traverse the vector
        let n = v.len();
        for i in 0..n - 1 {
            v[i].as_ref().unwrap().borrow_mut().left = None;
            v[i].as_ref().unwrap().borrow_mut().right = v[i + 1].clone();
        }
    }

    #[allow(dead_code)]
    fn pre_order_traverse(
        v: &mut Vec<Option<Rc<RefCell<TreeNode>>>>,
        root: &Option<Rc<RefCell<TreeNode>>>,
    ) {
        if root.is_none() {
            return;
        }
        v.push(root.clone());
        let left = root.as_ref().unwrap().borrow().left.clone();
        let right = root.as_ref().unwrap().borrow().right.clone();
        Self::pre_order_traverse(v, &left);
        Self::pre_order_traverse(v, &right);
    }
}

// Accepted solution for LeetCode #114: Flatten Binary Tree to Linked List
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

/**
 Do not return anything, modify root in-place instead.
 */
function flatten(root: TreeNode | null): void {
    while (root !== null) {
        if (root.left !== null) {
            let pre = root.left;
            while (pre.right !== null) {
                pre = pre.right;
            }
            pre.right = root.right;
            root.right = root.left;
            root.left = null;
        }
        root = root.right;
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.