# 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]

示例二:

输入:root = []
输出:[]

示例 3:

输入:root = [1]
输出:[1]

tips:

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

# 解法一:使用递归

记忆:中序遍历不忘 “左链入栈”

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

# 解法二:使用栈

class Solution {
  
    public List<Integer> inorderTraversal(TreeNode root) {
        	List<Integer> list = new ArrayList<>();
            Stack<TreeNode> stack = new Stack<>();
            TreeNode cur = root;
            while (cur != null || !stack.isEmpty()) {
                if (cur != null) {
                    stack.push(cur);
                    cur = cur.left;
                } else {
                    cur = stack.pop();
                    list.add(cur.val);
                    cur = cur.right;
                }
            }
            return list;
    }
}