feat: add solutions to lc problem: No.0623

No.0623.Add One Row to Tree

solution/0600-0699/0623.Add One Row to Tree/README.md

@@ -54,6 +54,10 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：DFS**
+
+**方法二：BFS**
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -68,24 +72,50 @@
 #         self.left = left
 #         self.right = right
 class Solution:
-    def addOneRow(self, root: TreeNode, val: int, depth: int) -> TreeNode:
+    def addOneRow(self, root: Optional[TreeNode], val: int, depth: int) -> Optional[TreeNode]:
         def dfs(root, d):
             if root is None:
                 return
             if d == depth - 1:
-                l = TreeNode(val=val, left=root.left)
-                r = TreeNode(val=val, right=root.right)
-                root.left, root.right = l, r
+                root.left = TreeNode(val, root.left, None)
+                root.right = TreeNode(val, None, root.right)
                 return
             dfs(root.left, d + 1)
             dfs(root.right, d + 1)
 
         if depth == 1:
-            return TreeNode(val=val, left=root)
+            return TreeNode(val, root)
         dfs(root, 1)
         return root
 ```
 
+```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 addOneRow(self, root: Optional[TreeNode], val: int, depth: int) -> Optional[TreeNode]:
+        if depth == 1:
+            return TreeNode(val, root)
+        q = deque([root])
+        i = 0
+        while q:
+            i += 1
+            for _ in range(len(q)):
+                node = q.popleft()
+                if node.left:
+                    q.append(node.left)
+                if node.right:
+                    q.append(node.right)
+                if i == depth - 1:
+                    node.left = TreeNode(val, node.left, None)
+                    node.right = TreeNode(val, None, node.right)
+        return root
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -137,6 +167,51 @@ class Solution {
 }
 ```
 
+```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 {
+    public TreeNode addOneRow(TreeNode root, int val, int depth) {
+        if (depth == 1) {
+            return new TreeNode(val, root, null);
+        }
+        Deque<TreeNode> q = new ArrayDeque<>();
+        q.offer(root);
+        int i = 0;
+        while (!q.isEmpty()) {
+            ++i;
+            for (int k = q.size(); k > 0; --k) {
+                TreeNode node = q.pollFirst();
+                if (node.left != null) {
+                    q.offer(node.left);
+                }
+                if (node.right != null) {
+                    q.offer(node.right);
+                }
+                if (i == depth - 1) {
+                    node.left = new TreeNode(val, node.left, null);
+                    node.right = new TreeNode(val, null, node.right);
+                }
+            }
+        }
+        return root;
+    }
+}
+```
+
 ### **C++**
 
 ```cpp
@@ -180,6 +255,45 @@ public:
 };
 ```
 
+```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* addOneRow(TreeNode* root, int val, int depth) {
+        if (depth == 1) return new TreeNode(val, root, nullptr);
+        queue<TreeNode*> q{{root}};
+        int i = 0;
+        while (!q.empty())
+        {
+            ++i;
+            for (int k = q.size(); k; --k)
+            {
+                TreeNode* node = q.front();
+                q.pop();
+                if (node->left) q.push(node->left);
+                if (node->right) q.push(node->right);
+                if (i == depth - 1)
+                {
+                    node->left = new TreeNode(val, node->left, nullptr);
+                    node->right = new TreeNode(val, nullptr, node->right);
+                }
+            }
+        }
+        return root;
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -213,8 +327,84 @@ func addOneRow(root *TreeNode, val int, depth int) *TreeNode {
 }
 ```
 
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func addOneRow(root *TreeNode, val int, depth int) *TreeNode {
+	if depth == 1 {
+		return &TreeNode{val, root, nil}
+	}
+	q := []*TreeNode{root}
+	i := 0
+	for len(q) > 0 {
+		i++
+		for k := len(q); k > 0; k-- {
+			node := q[0]
+			q = q[1:]
+			if node.Left != nil {
+				q = append(q, node.Left)
+			}
+			if node.Right != nil {
+				q = append(q, node.Right)
+			}
+			if i == depth-1 {
+				node.Left = &TreeNode{val, node.Left, nil}
+				node.Right = &TreeNode{val, nil, node.Right}
+			}
+		}
+	}
+	return root
+}
+```
+
 ### **TypeScript**
 
+```ts
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function addOneRow(
+    root: TreeNode | null,
+    val: number,
+    depth: number,
+): TreeNode | null {
+    function dfs(root, d) {
+        if (!root) {
+            return;
+        }
+        if (d == depth - 1) {
+            root.left = new TreeNode(val, root.left, null);
+            root.right = new TreeNode(val, null, root.right);
+            return;
+        }
+        dfs(root.left, d + 1);
+        dfs(root.right, d + 1);
+    }
+    if (depth == 1) {
+        return new TreeNode(val, root);
+    }
+    dfs(root, 1);
+    return root;
+}
+```
+
 ```ts
 /**
  * Definition for a binary tree node.