力扣：94:二叉树的中序遍历

LeetCode/src/main/java/leetcode/editor/cn/BinaryTreeInorderTraversal.java

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+//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]
+// 
+//
+// 示例 4： 
+//
+// 
+//输入：root = [1,2]
+//输出：[2,1]
+// 
+//
+// 示例 5： 
+//
+// 
+//输入：root = [1,null,2]
+//输出：[1,2]
+// 
+//
+// 
+//
+// 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？ 
+// Related Topics 栈 树 哈希表 
+// 👍 902 👎 0
+
+package leetcode.editor.cn;
+
+import com.code.leet.entiy.TreeNode;
+
+import java.util.ArrayList;
+import java.util.List;
+
+//94:二叉树的中序遍历
+public class BinaryTreeInorderTraversal {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new BinaryTreeInorderTraversal().new Solution();
+    }
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+
+    /**
+     * 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) {
+            List<Integer> list = new ArrayList<>();
+            inorder(root, list);
+            return list;
+        }
+
+        private void inorder(TreeNode root, List<Integer> list) {
+            if (root == null) {
+                return;
+            }
+            inorder(root.left, list);
+            list.add(root.val);
+            inorder(root.right, list);
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

LeetCode/src/main/java/leetcode/editor/cn/BinaryTreeInorderTraversal.md

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+<p>给定一个二叉树的根节点 <code>root</code> ，返回它的 <strong>中序</strong> 遍历。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/09/15/inorder_1.jpg" style="width: 202px; height: 324px;" />
+<pre>
+<strong>输入：</strong>root = [1,null,2,3]
+<strong>输出：</strong>[1,3,2]
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>root = []
+<strong>输出：</strong>[]
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>root = [1]
+<strong>输出：</strong>[1]
+</pre>
+
+<p><strong>示例 4：</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/09/15/inorder_5.jpg" style="width: 202px; height: 202px;" />
+<pre>
+<strong>输入：</strong>root = [1,2]
+<strong>输出：</strong>[2,1]
+</pre>
+
+<p><strong>示例 5：</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/09/15/inorder_4.jpg" style="width: 202px; height: 202px;" />
+<pre>
+<strong>输入：</strong>root = [1,null,2]
+<strong>输出：</strong>[1,2]
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li>树中节点数目在范围 <code>[0, 100]</code> 内</li>
+	<li><code>-100 <= Node.val <= 100</code></li>
+</ul>
+
+<p> </p>
+
+<p><strong>进阶:</strong> 递归算法很简单，你可以通过迭代算法完成吗？</p>
+<div><div>Related Topics</div><div><li>栈</li><li>树</li><li>哈希表</li></div></div>\n<div><li>👍 902</li><li>👎 0</li></div>