package com.fishercoder.solutions;


import com.fishercoder.common.classes.TreeNode;

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

/**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?*/

public class _94 {

    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList();
        return inorder(root, result);
    }
    
    List<Integer> inorder(TreeNode root, List<Integer> result){
        if(root == null) return result;
        if(root.left != null) inorder(root.left, result);
        result.add(root.val);
        if(root.right != null) inorder(root.right, result);
        return result;
    }

    public List<Integer> inorderTraversal_iterative(TreeNode root) {
        List<Integer> result = new ArrayList();
        Stack<TreeNode> stack = new Stack();
        while(root != null || !stack.isEmpty()){
            while(root != null){
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            result.add(root.val);
            root = root.right;
        }
        return result;
    }

}

@@ -158,16 +158,60 @@
 ck;%0A%0A/**
+%0A * 94. Binary Tree Inorder Traversal%0A *%0A * 
 Given a 

edit 94