lc-94

//Given the root of a binary tree, return the inorder traversal of its nodes' va
//lues. 
//
// 
// Example 1: 
//
// 
//Input: root = [1,null,2,3]
//Output: [1,3,2]
// 
//
// Example 2: 
//
// 
//Input: root = []
//Output: []
// 
//
// Example 3: 
//
// 
//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? Related
// Topics Stack Tree Depth-First Search Binary Tree 
// 👍 8683 👎 405


//leetcode submit region begin(Prohibit modification and deletion)

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.Stack;

/**
 * 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) {
        Stack<TreeNode> nodeStack = new Stack<>();
        List<Integer> result = new ArrayList<>();
        Set<TreeNode> leftVisited = new HashSet<>();
        if (root == null) {
            return result;
        }

        nodeStack.push(root);

        while (!nodeStack.isEmpty()) {
            if (nodeStack.peek().left != null && !leftVisited.contains(nodeStack.peek().left)) {
                nodeStack.push(nodeStack.peek().left);
                leftVisited.add(nodeStack.peek());
            } else {
                TreeNode node = nodeStack.pop();
                result.add(node.val);

                if (node.right != null) {
                    nodeStack.push(node.right);
                }
            }
        }

        return result;
    }
}
//leetcode submit region end(Prohibit modification and deletion)