[LC] 94. Binary Tree Inorder Traversal

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

Example:

Input: [1,null,2,3]
   1
    
     2
    /
   3

Output: [1,3,2]

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        if root is None:
            return res
        stack = []
        cur = root
        while cur or stack:
            while cur:
                stack.append(cur)
                cur = cur.left
            node = stack.pop()
            res.append(node.val)
            cur = node.right
        return res
                

            
        

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        LinkedList<TreeNode> stack = new LinkedList<>();
        List<Integer> res = new ArrayList<>();
        TreeNode cur = root;
        while (!stack.isEmpty() || cur != null) {
            if (cur != null) {
                stack.offerFirst(cur);
                cur = cur.left;
            } else {
                cur = stack.pollFirst();
                res.add(cur.val);
                cur = cur.right;
            }
        }
        return res;
    }
}