In-order traversal of binary tree-LeetCode

topic:

Original link: https://leetcode-cn.com/problems/binary-tree-inorder-traversal/

Ideas:

1. Recursion or use the stack to simulate recursion
2. Time complexity is O(n)

Code (recursive):

/**
 * 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;
 *     }
 * }
 */
import java.util.ArrayList;
class Solution {
    
    
    public List<Integer> inorderTraversal(TreeNode root) {
    
    
        List<Integer> result = new ArrayList<>();
        if(root==null) return result;
        inorder(root, result);
        return result;
    }

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

Code (stack):

import java.util.ArrayList;
import java.util.Deque;
import java.util.LinkedList;
class Solution {
    
    
    public List<Integer> inorderTraversal(TreeNode root) {
    
    
        List<Integer> result = new ArrayList<>();
        Deque<TreeNode> stack = new LinkedList<>();
        while(root!=null || !stack.isEmpty()){
    
    
            // 深度添加左结点
            while(root!=null){
    
    
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            result.add(root.val);
            root = root.right;
        }
        return result;
    }
}