feat: add the solution of Second Minimum Node In a Binary Tree(671) with Java.

Author: Mukyu

README.md

@@ -69,7 +69,7 @@
 | [661][661-question] | [Image Smoother][661-tips]                                   | [✅][661-java] | [✅][661-js] |                  |
 | [665][665-question] | [Non-decreasing Array][665-tips]                             | [✅][665-java] | [✅][665-js] |                  |
 | [669][669-question] | [Trim a Binary Search Tree][669-tips]                        | [✅][669-java] | [✅][669-js] |                  |
-| [671][671-question] | [Second Minimum Node In a Binary Tree][671-tips]             |                | [✅][671-js] |                  |
+| [671][671-question] | [Second Minimum Node In a Binary Tree][671-tips]             | [✅][671-java] | [✅][671-js] |                  |
 | [674][674-question] | [Longest Continuous Increasing Subsequence][674-tips]        |                | [✅][674-js] |                  |
 | [680][680-question] | [Valid Palindrome II][680-tips]                              |                | [✅][680-js] |                  |
 | [682][682-question] | [Baseball Game][682-tips]                                    |                | [✅][682-js] |                  |
@@ -462,6 +462,7 @@
 [661-java]: ./src/_661/Solution.java
 [665-java]: ./src/_665/Solution.java
 [669-java]: ./src/_669/Solution.java
+[671-java]: ./src/_671/Solution.java
 [728-java]: ./src/_728/Solution.java
 [771-java]: ./src/_771/Solution.java
 [804-java]: ./src/_804/Solution.java

src/_671/Solution.java

@@ -0,0 +1,49 @@
+package _671;
+
+import java.util.Stack;
+
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public int findSecondMinimumValue(TreeNode root) {
+        Stack<TreeNode> stack = new Stack<>();
+        int min = root.val;
+        int secMin = -1;
+        stack.push(root);
+
+        while (!stack.empty()) {
+            TreeNode node = stack.pop();
+
+            if (node.val > min) {
+                 if (node.val < secMin || secMin == -1) {
+                     secMin = node.val;
+                 }
+                 continue;
+            }
+
+            if (node.left != null) {
+                stack.push(node.right);
+                stack.push(node.left);
+            }
+        }
+
+        return secMin;
+    }
+
+    class TreeNode {
+        int val;
+        TreeNode left;
+        TreeNode right;
+
+        TreeNode(int x) {
+            val = x;
+        }
+    }
+}

tips/671/README.md

@@ -1,46 +1,76 @@
-[xxxx][title]
+[Second Minimum Node In a Binary Tree][title]
 
 ## Description
-// 抄题目
+Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly `two` or `zero` sub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes.
 
+Given such a binary tree, you need to output the **second minimum** value in the set made of all the nodes' value in the whole tree.
 
-**Example:**
+If no such second minimum value exists, output -1 instead.
+
+**Example 1:**
 
 ```
-// 抄Example
+Input: 
+    2
+   / \
+  2   5
+     / \
+    5   7
+
+Output: 5
+Explanation: The smallest value is 2, the second smallest value is 5.
 ```
 
-**Note:**
-// Note
-
-**Tags:** // tags
-
-
-## 思路 1
-// 贴一些关键代码,说一些解题思路
-// (同一种语言可以写多种思路，与某种语言思路相同的另一种语言的思路无须赘述，但可以把代码贴在后面) 
-```java
+**Example 2:**
 
 ```
-```javascript
+Input: 
+    2
+   / \
+  2   2
 
+Output: -1
+Explanation: The smallest value is 2, but there isn't any second smallest value.
 ```
 
-## 思路 2
-// 贴一些关键代码,说一些解题思路
-```java
+**Tags:** [Tree](https://leetcode.com/tag/tree/)
 
-```
+## 思路
+
+题目给出一个特殊的二叉树：任意节点，要么没有子节点，要么有两个子节点，且子节点的值会大于等于他们的根节点。要求找出第二小的节点值。根据此二叉树的特性，根节点的值为最小值。先对树进行先序遍历，当发现比根节点值（最小值）大的节点后，后面的子树就可以进行剪枝了，因为后面的子节点都大于或等于此节点，不会是第二小的节点（或者与第二小的节点值相等）。
 
-## 思路 3
-// 贴一些关键代码,说一些解题思路
-```kotlin
+**Java:**
 
+```java
+public int findSecondMinimumValue(TreeNode root) {
+    Stack<TreeNode> stack = new Stack<>();
+    int min = root.val;
+    int secMin = -1;
+    stack.push(root);
+
+    while (!stack.empty()) {
+        TreeNode node = stack.pop();
+
+        if (node.val > min) {
+             if (node.val < secMin || secMin == -1) {
+                 secMin = node.val;
+             }
+             continue;
+        }
+
+        if (node.left != null) {
+            stack.push(node.right);
+            stack.push(node.left);
+        }
+    }
+
+    return secMin;
+}
 ```
 
 ## 结语
-   
+
 如果你同我们一样热爱数据结构、算法、LeetCode，可以关注我们 GitHub 上的 LeetCode 题解：[LeetCode-Solution][ls]
 
-[title]: https://leetcode.com/problems/xxxx
+[title]: https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/description/
 [ls]: https://github.com/SDE603/LeetCode-Solution