Binary Tree Inorder Traversal (Iterative)

Easy Trees & BST Recursion Original on LeetCode

Binary Tree Inorder Traversal (Iterative) is a easy trees & bst problem solved with the 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 (reference) up.

Problem

Return the inorder traversal of a binary tree iteratively, using your own stack in place of recursion.

Examples

Example 1

Input: root = [5, 3, 8, 1, 4]
Output: [1, 3, 4, 5, 8]

Example 2

Input: root = []
Output: []

Constraints

The tree has 0 to 100 nodes.
Solve it without recursion.

Animated walkthrough

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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 (reference)	`O(n)`	`O(h)`
Iterative (explicit stack)	`O(n)`	`O(h)`

1Recursive (reference)

TimeO(n)

SpaceO(h)

Left, node, right with the call stack.

inorder(left); add; inorder(right).

Java

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

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

2Iterative (explicit stack)

TimeO(n)

SpaceO(h)

The stack holds the nodes whose left subtree is being explored. Push down the left spine; pop gives the next node in sorted order; then continue with its right subtree.

cur = root.
While cur != null or the stack is not empty: while cur != null: push, cur = cur.left.
cur = pop(); visit; cur = 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

Only right children
Only left children

Hints

Hint 1

Keep going left, pushing nodes. When you cannot go further, pop, visit, then turn right.

FAQ

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

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

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

It is a trees & bst problem that uses the recursion pattern. Other problems with the same pattern: Binary Tree Inorder Traversal (Recursive), Binary Tree Preorder Traversal (Recursive), Binary Tree Postorder Traversal (Recursive).

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

Yes. Recursive (reference) takes O(n) time and O(h) space. Left, node, right with the call stack.

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

Only right children; Only left children.