Binary Tree Inorder Traversal (Recursive) (Recursion & Backtracking)

Easy Recursion & Backtracking Basic Recursion Original on LeetCode

Binary Tree Inorder Traversal (Recursive) is a easy recursion & backtracking problem solved with the basic recursion pattern. The best approach, iterative (explicit stack), runs in O(n) time and O(h) space. Below are 2 approaches in Java, from recursive up.

Problem

Return the inorder traversal of a binary tree: left subtree, node, right subtree. You already did this in class; here it is the warm-up for thinking recursively about trees.

Examples

Example 1

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

Example 2

Input: root = [2, 1, 3]
Output: [1, 2, 3]
Why: Inorder on a BST gives sorted order.

Constraints

The tree has 0 to 100 nodes.

Animated walkthrough

A narrated, step-by-step animation that builds the solution from the idea up. Press play, or step through it at your own pace.

Concepts first, then the problem and every approach, step by step.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Recursive	`O(n)`	`O(h)`
Iterative (explicit stack)	`O(n)`	`O(h)`

1Recursive

TimeO(n)

SpaceO(h)h is the tree height: O(log n) if balanced, O(n) if skewed.

Left subtree, then the node, then the right subtree. The recursion is the problem's definition, which is why this is the classic first tree-recursion exercise.

If node is null, return.
go(left); add node.val; go(right).

Java

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> out = new ArrayList<>();
        go(root, out);
        return out;
    }

    private void go(TreeNode node, List<Integer> out) {
        if (node == null) return;
        go(node.left, out);
        out.add(node.val);
        go(node.right, out);
    }
}

2Iterative (explicit stack)

TimeO(n)

SpaceO(h)

Simulate the call stack: push nodes while going left, then pop one, visit it and switch to its right child.

While cur != null or the stack is not empty: push all lefts; pop; visit; cur = right.

Java

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> out = new ArrayList<>();
        Deque<TreeNode> st = new ArrayDeque<>();
        TreeNode cur = root;
        while (cur != null || !st.isEmpty()) {
            while (cur != null) { st.push(cur); cur = cur.left; }
            cur = st.pop();
            out.add(cur.val);
            cur = cur.right;
        }
        return out;
    }
}

Edge cases to test

Empty tree
Skewed tree (depth n)

Hints

Hint 1

Trust the recursion: assume inorder(left) already works, then add the root, then inorder(right).

FAQ

What is the best time complexity for Binary Tree Inorder Traversal (Recursive)?

Iterative (explicit stack) runs in O(n) time and O(h) extra space.

Which pattern does Binary Tree Inorder Traversal (Recursive) use?

It is a recursion & backtracking problem that uses the basic recursion pattern. Other problems with the same pattern: Factorial of a number, Fibonacci Number, Basic Backtracking Template.

Is there a brute force solution for Binary Tree Inorder Traversal (Recursive)?

Yes. Recursive takes O(n) time and O(h) space. Left subtree, then the node, then the right subtree.

Which edge cases should I test for Binary Tree Inorder Traversal (Recursive)?

Empty tree; Skewed tree (depth n).