Binary Tree Inorder Traversal

LeetCode 94 | Difficulty: Easy

Easy

Problem Description

Given the root of a binary tree, return the inorder traversal of its nodes' values.

Example 1:

Input: root = [1,null,2,3]

Output: [1,3,2]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [4,2,6,5,7,1,3,9,8]

Explanation:

Example 3:

Input: root = []

Output: []

Example 4:

Input: root = [1]

Output: [1]

Constraints:

- The number of nodes in the tree is in the range `[0, 100]`.

- `-100 <= Node.val <= 100`

Follow up: Recursive solution is trivial, could you do it iteratively?

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

---

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.

---

Solutions

Solution 1: C# (Best: 173 ms)

Metric	Value
Runtime	173 ms
Memory	39.1 MB
Date	2021-12-08

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 IList<int> InorderTraversal(TreeNode root) {
        List<int> result = new List<int>();
    TreeNode current = root, prev = null;
    
    while(current != null)
    {
        if(current.left == null)
        {
            result.Add(current.val);
            current = current.right;
        }
        else
        {
            // find predecessor
            prev = current.left;
            while(prev.right != null && prev.right != current)
            {
                prev = prev.right;
            }
            if(prev.right == null)
            {
                prev.right = current;
                current = current.left;
            }
            else
            {
                prev.right = null;
                result.Add(current.val);
                current = current.right;
            }
        }
    }
    return result;
    }
}

📜 3 more C# submission(s)

Submission (2019-12-31) — 240 ms, 29.3 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public IList<int> InorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
    List<int> inorder = new List<int>();
    var current = root;
    if(root==null) return inorder;
    while(stack.Count!=0 || current != null )
    {
        while(current != null)
        {
            stack.Push(current);
            current = current.left;
        }
        var popped = stack.Pop();
        inorder.Add(popped.val);
        current = popped.right;
    }
        
    
    
    return inorder;
    }
}

Submission (2018-05-02) — 280 ms, N/A

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public IList<int> InorderTraversal(TreeNode root) {
        System.Collections.Generic.Stack<TreeNode> stack = new Stack<TreeNode>();
    List<int> inorder = new List<int>();
    var current = root;
    if(root==null) return inorder;
    while(stack.Count!=0 || current != null )
    {
        if(current != null)
        {
            stack.Push(current);
            current = current.left;
        }
        else
        {
            if(stack.Count!=0)
            {
                var popped = stack.Pop();
                inorder.Add(popped.val);
                current = popped.right;
            }
        }
    }
    
    return inorder;
    }
}

Submission (2018-01-11) — 601 ms, N/A

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public IList<int> InorderTraversal(TreeNode root) {
        System.Collections.Generic.Stack<TreeNode> stack = new Stack<TreeNode>();
    List<int> inorder = new List<int>();
    var current = root;
    while(stack.Count!=0 || current != null )
    {
        if(current != null)
        {
            stack.Push(current);
            current = current.left;
        }
        else
        {
            if(stack.Count!=0)
            {
                var popped = stack.Pop();
                inorder.Add(popped.val);
                current = popped.right;
            }
        }
    }
    
    return inorder;
    }
}

---

Complexity Analysis

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

---

Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.
• Think about what triggers a pop: is it finding a match, or finding a smaller/larger element?