[LeetCode] 94. Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree [1,null,2,3],
   1
    \
     2
    /
   3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?

Thought process:
Inorder traversal visits a tree in the order of left subtree - root - right subtree.

Solution 1 (Recursive):
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> inorder = new ArrayList<>();
        
        inorderTraversal(root, inorder);
        
        return inorder;
    }
    
    private void inorderTraversal(TreeNode root, List<Integer> inorder) {
        if (root == null) {
            return;
        }
        
        inorderTraversal(root.left, inorder);
        inorder.add(root.val);
        inorderTraversal(root.right, inorder);
    }
}

Time complexity: O(n).

The recursive solution is trivial. To solve this iteratively, we need to use a stack to simulate the function call stack. To achieve inorder traversal, we go to the left subtree until the leaf, and add nodes along the way.

Solution 2 (Iterative):
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> inorder = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();

        while (root != null || !stack.isEmpty()) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            
            root = stack.pop();
            inorder.add(root.val);
            root = root.right;
        }
        
        return inorder;
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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