LeetCode 94. 二叉树的中序遍历 Binary Tree Inorder Traversal

题目: LeetCode 94. 二叉树的中序遍历

给定一个二叉树，返回它的中序 遍历。

示例:

输入: [1,null,2,3]
1

2
/
3

输出: [1,3,2]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？

递归算法

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

/// 94. Binary Tree Inorder Traversal
/// https://leetcode.com/problems/binary-tree-inorder-traversal/solution/
/// 二叉树的中序遍历
/// 时间复杂度: O(n), n为树的节点个数
/// 空间复杂度: O(h), h为树的高度
public class Solution094 {

    // Definition for a binary tree node.
    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
        TreeNode(int x) { val = x; }
    }

    public List<Integer> inorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        inorderTraversal(root, res);
        return res;
    }

    private void inorderTraversal(TreeNode node, List<Integer> list){
        if(node != null){
            inorderTraversal(node.left, list);
            list.add(node.val);
            inorderTraversal(node.right, list);
        }
    }
}

非递归算法

// Definition for a binary tree node.
public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

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

// Classic Non-Recursive algorithm for inorder traversal
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
public class Solution1 {

    public List<Integer> inorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;

        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while(cur != null || !stack.empty()){

            while(cur != null){
                stack.push(cur);
                cur = cur.left;
            }

            cur = stack.pop();
            res.add(cur.val);
            cur = cur.right;
        }
        return res;
    }
}

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

// Another Classic Non-Recursive algorithm for inorder traversal
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
public class Solution2 {

    public List<Integer> inorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;

        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while(cur != null || !stack.empty()){

            if(cur != null){
                stack.push(cur);
                cur = cur.left;
            }
            else{
                cur = stack.pop();
                res.add(cur.val);
                cur = cur.right;
            }
        }
        return res;
    }
}