📝 add lc 94

DreamCats · DreamCats · commit eb95bf784f22 · 2021-04-15T14:38:05.000+08:00
diff --git a/Java/alg/README.md b/Java/alg/README.md
@@ -133,7 +133,8 @@
 - [91.解码方法](lc/91.解码方法.md)
 - [92.反转链表2](92.反转链表2.md)
 - [93.复原IP地址](93.复原IP地址.md)
+- [94.二叉树的中序遍历](94.二叉树的中序遍历.md)
 
 ## 笔试
 
-<!-- - [1.分苹果](we/1.分苹果.md) -->
+<!-- - [1.分苹果](we/1.分苹果.md) -->
diff --git a/Java/alg/lc/94.二叉树的中序遍历.md b/Java/alg/lc/94.二叉树的中序遍历.md
@@ -0,0 +1,93 @@
+# 94.二叉树的中序遍历
+
+[url](https://leetcode-cn.com/problems/binary-tree-inorder-traversal/)
+
+## 题目
+
+给定一个二叉树的根节点 `root` ，返回它的 **中序** 遍历。
+
+![](https://assets.leetcode.com/uploads/2020/09/15/inorder_1.jpg)
+
+
+```
+输入：root = [1,null,2,3]
+输出：[1,3,2]
+```
+
+## 方法
+
+
+## code
+
+### js
+
+```js
+var inorderTraversal = function(root) {
+    let ret = [];
+    if (root === null)
+        return ret;
+    let stack = [];
+    let cur = root;
+    while (cur !== null || stack.length !== 0) {
+        while (cur !== null) {
+            stack.push(cur);
+            cur = cur.left;
+        }
+        let t = stack.pop();
+        ret.push(t.val)
+        cur = t.right;
+    }
+    return ret;
+};
+```
+
+### go
+
+```go
+func inorderTraversal(root *TreeNode) []int {
+	var ret []int
+	if root == nil {
+		return ret
+	}
+	var stack []*TreeNode
+	cur := root
+	for cur != nil || len(stack) != 0 {
+		for cur != nil {
+			stack = append(stack, cur)
+			cur = cur.Left
+		}
+		t := stack[len(stack)-1]
+		stack = stack[:len(stack)-1]
+		ret = append(ret, t.Val)
+		cur = t.Right
+	}
+	return ret
+}
+```
+
+### java
+
+```java
+class Solution {
+    public List<Integer> inorderTraversal(TreeNode root) {
+        List<Integer> ret = new ArrayList<>();
+        if (root == null)
+            return ret;
+        Stack<TreeNode> stack = new Stack<>();
+
+        TreeNode cur = root;
+
+        while (cur != null || !stack.isEmpty()) {
+            while (cur != null) {
+                stack.push(cur);
+                cur = cur.left;
+            }
+            TreeNode t = stack.pop();
+            ret.add(t.val);
+            cur = t.right;
+        }
+        return ret;
+    }
+}
+```
+
