0094 - Binary Tree Inorder Traversal

0094 - Binary Tree Inorder Traversal

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

Examples

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

Input: root = [] Output: []

Input: root = [1] Output: [1]

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

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

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?

Java Solution

/**
 * 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 List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> inorder = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode pointer = root;

        while(pointer != null) {
            stack.push(pointer);
            pointer = pointer.left;
        }

        while(!stack.isEmpty()) {
            TreeNode t = stack.pop();
            inorder.add(t.val);

            t = t.right;
            while(t != null) {
                stack.push(t);
                t = t.left;
            }
        }

        return inorder;
    }
}