feat: add solutions to lc problem: No.1315. Sum of Nodes with Even-Valued Grandparent

Author: yanglbme

solution/1300-1399/1315.Sum of Nodes with Even-Valued Grandparent/README.md

@@ -34,27 +34,167 @@
 	<li>每个节点的值在&nbsp;<code>1</code> 到&nbsp;<code>100</code> 之间。</li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+深度优先搜索。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def sumEvenGrandparent(self, root: TreeNode) -> int:
+        self.res = 0
+
+        def dfs(g, p):
+            if p is None:
+                return
+            if g.val % 2 == 0:
+                if p.left:
+                    self.res += p.left.val
+                if p.right:
+                    self.res += p.right.val
+            dfs(p, p.left)
+            dfs(p, p.right)
+
+        dfs(root, root.left)
+        dfs(root, root.right)
+        return self.res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```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 {
+    private int res;
+
+    public int sumEvenGrandparent(TreeNode root) {
+        res = 0;
+        dfs(root, root.left);
+        dfs(root, root.right);
+        return res;
+    }
+
+    private void dfs(TreeNode g, TreeNode p) {
+        if (p == null) {
+            return;
+        }
+        if (g.val % 2 == 0) {
+            if (p.left != null) {
+                res += p.left.val;
+            }
+            if (p.right != null) {
+                res += p.right.val;
+            }
+        }
+        dfs(p, p.left);
+        dfs(p, p.right);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    int res;
+
+    int sumEvenGrandparent(TreeNode* root) {
+        res = 0;
+        dfs(root, root->left);
+        dfs(root, root->right);
+        return res;
+    }
+
+    void dfs(TreeNode* g, TreeNode* p) {
+        if (!p) return;
+        if (g->val % 2 == 0)
+        {
+            if (p->left) res += p->left->val;
+            if (p->right) res += p->right->val;
+        }
+        dfs(p, p->left);
+        dfs(p, p->right);
+    }
+};
+```
 
+### **Go**
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+var res int
+
+func sumEvenGrandparent(root *TreeNode) int {
+	res = 0
+	dfs(root, root.Left)
+	dfs(root, root.Right)
+	return res
+}
+
+func dfs(g, p *TreeNode) {
+	if p == nil {
+		return
+	}
+	if g.Val%2 == 0 {
+		if p.Left != nil {
+			res += p.Left.Val
+		}
+		if p.Right != nil {
+			res += p.Right.Val
+		}
+	}
+	dfs(p, p.Left)
+	dfs(p, p.Right)
+}
 ```
 
 ### **...**

solution/1300-1399/1315.Sum of Nodes with Even-Valued Grandparent/README_EN.md

@@ -6,22 +6,14 @@
 
 <p>Given a binary tree, return the sum of values of nodes with even-valued grandparent.&nbsp; (A <em>grandparent</em> of a node is the parent of its parent, if it exists.)</p>
 
-
-
 <p>If there are no nodes with an even-valued grandparent, return&nbsp;<code>0</code>.</p>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Example 1:</strong></p>
 
-
-
 <p><strong><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1315.Sum%20of%20Nodes%20with%20Even-Valued%20Grandparent/images/1473_ex1.png" style="width: 350px; height: 214px;" /></strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
@@ -32,33 +24,170 @@
 
 </pre>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Constraints:</strong></p>
 
-
-
 <ul>
 	<li>The number of nodes in the tree is between&nbsp;<code>1</code>&nbsp;and&nbsp;<code>10^4</code>.</li>
 	<li>The value of nodes is between&nbsp;<code>1</code>&nbsp;and&nbsp;<code>100</code>.</li>
 </ul>
 
 ## Solutions
 
+DFS.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def sumEvenGrandparent(self, root: TreeNode) -> int:
+        self.res = 0
+
+        def dfs(g, p):
+            if p is None:
+                return
+            if g.val % 2 == 0:
+                if p.left:
+                    self.res += p.left.val
+                if p.right:
+                    self.res += p.right.val
+            dfs(p, p.left)
+            dfs(p, p.right)
+
+        dfs(root, root.left)
+        dfs(root, root.right)
+        return self.res
 ```
 
 ### **Java**
 
 ```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 {
+    private int res;
+
+    public int sumEvenGrandparent(TreeNode root) {
+        res = 0;
+        dfs(root, root.left);
+        dfs(root, root.right);
+        return res;
+    }
+
+    private void dfs(TreeNode g, TreeNode p) {
+        if (p == null) {
+            return;
+        }
+        if (g.val % 2 == 0) {
+            if (p.left != null) {
+                res += p.left.val;
+            }
+            if (p.right != null) {
+                res += p.right.val;
+            }
+        }
+        dfs(p, p.left);
+        dfs(p, p.right);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    int res;
+
+    int sumEvenGrandparent(TreeNode* root) {
+        res = 0;
+        dfs(root, root->left);
+        dfs(root, root->right);
+        return res;
+    }
+
+    void dfs(TreeNode* g, TreeNode* p) {
+        if (!p) return;
+        if (g->val % 2 == 0)
+        {
+            if (p->left) res += p->left->val;
+            if (p->right) res += p->right->val;
+        }
+        dfs(p, p->left);
+        dfs(p, p->right);
+    }
+};
+```
 
+### **Go**
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+var res int
+
+func sumEvenGrandparent(root *TreeNode) int {
+	res = 0
+	dfs(root, root.Left)
+	dfs(root, root.Right)
+	return res
+}
+
+func dfs(g, p *TreeNode) {
+	if p == nil {
+		return
+	}
+	if g.Val%2 == 0 {
+		if p.Left != nil {
+			res += p.Left.Val
+		}
+		if p.Right != nil {
+			res += p.Right.Val
+		}
+	}
+	dfs(p, p.Left)
+	dfs(p, p.Right)
+}
 ```
 
 ### **...**