🔄Update: 二叉树的中序遍历

二叉树/1. 二叉树的中序遍历.md

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+# 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;
+    }
+}
+```