feat: add solutions to lc problem: No.0993

No.0993.Cousins in Binary Tree

Author: yanglbme

solution/0900-0999/0993.Cousins in Binary Tree/README.md

@@ -55,14 +55,20 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-BFS 实现。可以利用数组 p, d 记录每个节点对应的父节点以及深度。
+**1. BFS 实现**
+
+可以利用数组 p, d 记录每个节点对应的父节点以及深度。
+
+**2. DFS 实现**
 
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
+BFS：
+
 ```python
 # Definition for a binary tree node.
 # class TreeNode:
@@ -91,10 +97,40 @@ class Solution:
         return p[x] != p[y] and d[x] == d[y]
 ```
 
+DFS：
+
+```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 isCousins(self, root: TreeNode, x: int, y: int) -> bool:
+        p1 = p2 = d1 = d2 = None
+
+        def dfs(root, p, d):
+            if root is None:
+                return
+            nonlocal p1, p2, d1, d2, x, y
+            if root.val == x:
+                p1, d1 = p, d
+            if root.val == y:
+                p2, d2 = p, d
+            dfs(root.left, root, d + 1)
+            dfs(root.right, root, d + 1)
+
+        dfs(root, None, 0)
+        return p1 != p2 and d1 == d2
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
+BFS：
+
 ```java
 /**
  * Definition for a binary tree node.
@@ -139,8 +175,58 @@ class Solution {
 }
 ```
 
+DFS：
+
+```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 x, y;
+    private TreeNode p1, p2;
+    private int d1, d2;
+
+    public boolean isCousins(TreeNode root, int x, int y) {
+        this.x = x;
+        this.y = y;
+        dfs(root, null, 0);
+        return p1 != p2 && d1 == d2;
+    }
+
+    private void dfs(TreeNode root, TreeNode p, int d) {
+        if (root == null) {
+            return;
+        }
+        if (root.val == x) {
+            p1 = p;
+            d1 = d;
+        }
+        if (root.val == y) {
+            p2 = p;
+            d2 = d;
+        }
+        dfs(root.left, root, d + 1);
+        dfs(root.right, root, d + 1);
+    }
+}
+```
+
 ### **C++**
 
+BFS：
+
 ```cpp
 /**
  * Definition for a binary tree node->
@@ -187,8 +273,56 @@ public:
 };
 ```
 
+DFS：
+
+```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:
+    TreeNode* p1;
+    TreeNode* p2;
+    int d1, d2;
+    int x, y;
+
+    bool isCousins(TreeNode* root, int x, int y) {
+        this->x = x;
+        this->y = y;
+        dfs(root, nullptr, 0);
+        return p1 != p2 && d1 == d2;
+    }
+
+    void dfs(TreeNode* root, TreeNode* p, int d) {
+        if (!root) return;
+        if (root->val == x)
+        {
+            p1 = p;
+            d1 = d;
+        }
+        if (root->val == y)
+        {
+            p2 = p;
+            d2 = d;
+        }
+        dfs(root->left, root, d + 1);
+        dfs(root->right, root, d + 1);
+    }
+};
+```
+
 ### **Go**
 
+BFS：
+
 ```go
 /**
  * Definition for a binary tree node.
@@ -226,6 +360,44 @@ func isCousins(root *TreeNode, x int, y int) bool {
 }
 ```
 
+DFS：
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+var p1 *TreeNode
+var p2 *TreeNode
+var d1 int
+var d2 int
+
+func isCousins(root *TreeNode, x int, y int) bool {
+	dfs(root, nil, x, y, 0)
+	return p1 != p2 && d1 == d2
+}
+
+func dfs(root, p *TreeNode, x, y, d int) {
+	if root == nil {
+		return
+	}
+	if root.Val == x {
+		p1 = p
+		d1 = d
+	}
+	if root.Val == y {
+		p2 = p
+		d2 = d
+	}
+	dfs(root.Left, root, x, y, d+1)
+	dfs(root.Right, root, x, y, d+1)
+}
+```
+
 ### **...**
 
 ```