LeetCode/二叉树/1. 二叉树的中序遍历.md

# 二叉树的中序遍历

## 题目
给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。

 

示例 1：


输入：root = [1,null,2,3]
输出：[1,3,2]
示例 2：

输入：root = []
输出：[]
示例 3：

输入：root = [1]
输出：[1]
 

提示：

树中节点数目在范围 [0, 100] 内
-100 <= Node.val <= 100
 

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

## 思路

- 递归
- 迭代

## 代码

- 递归

```java
/**
 * 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;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        // Init: Res
        List<Integer> res = new ArrayList<>();
        inorder(root, res);
        return res;
    }

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

- 迭代

```java
/**
 * 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;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        // Init: stack, res
        Deque<TreeNode> stack = new LinkedList<>();
        List<Integer> res = new ArrayList<>();
        TreeNode cur = root;
        // Traverse 1: Judge
        while (cur != null || stack.size() > 0) {
            // Traverse 2: Add
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.peek();
            stack.pop();
            res.add(cur.val);
            cur = cur.right;
        }
        return res;
    }
}
```