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/*
 * @lc app=leetcode id=94 lang=java
 *
 * [94] Binary Tree Inorder Traversal
 *
 * https://leetcode.com/problems/binary-tree-inorder-traversal/description/
 *
 * algorithms
 * Easy (76.08%)
 * Likes:    13350
 * Dislikes: 776
 * Total Accepted:    2.5M
 * Total Submissions: 3.3M
 * Testcase Example:  '[1,null,2,3]'
 *
 * Given the root of a binary tree, return the inorder traversal of its nodes'
 * values.
 * 
 * 
 * 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?
 */

// @lc code=start

import java.util.ArrayList;

/**
 * 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;
 *     }
 * }
 */

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 /*
 * 思路：遞歸實現中序遍歷
 * */
class Solution1 {
    List<Integer> result = new ArrayList<>();
    public List<Integer> inorderTraversal(TreeNode root) {
        
        if (null == root) {

            return result;
        }
        if(root.left != null){
            inorderTraversal(root.left);
        }
        result.add(root.val);
        if(root.right != null){
            inorderTraversal(root.right);
        }
        return result;
    }
    
}

 /*
 * 思路：迭代實現中序遍歷
 * */
class Solution {
    List<Integer> result = new ArrayList<>();
    public List<Integer> inorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while(cur != null || !stack.isEmpty()){
            while(cur!= null){
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            result.add(cur.val);
            cur = cur.right;
        }
        return result;
    }
    
}