feat: add solutions to lc problem: No.0102

No.0102.Binary Tree Level Order Traversal

Author: yanglbme

solution/0100-0199/0102.Binary Tree Level Order Traversal/README.md

@@ -46,6 +46,15 @@
 
 **方法一：BFS**
 
+我们可以使用 BFS 的方法来解决这道题。首先将根节点入队，然后不断地进行以下操作，直到队列为空：
+
+-   遍历当前队列中的所有节点，将它们的值存储到一个临时数组 $t$ 中，然后将它们的孩子节点入队。
+-   将临时数组 $t$ 存储到答案数组中。
+
+最后返回答案数组即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点个数。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -60,10 +69,10 @@
 #         self.left = left
 #         self.right = right
 class Solution:
-    def levelOrder(self, root: TreeNode) -> List[List[int]]:
-        if root is None:
-            return []
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
         ans = []
+        if root is None:
+            return ans
         q = deque([root])
         while q:
             t = []
@@ -172,14 +181,13 @@ public:
  *     Right *TreeNode
  * }
  */
-func levelOrder(root *TreeNode) [][]int {
-	var ans [][]int
+func levelOrder(root *TreeNode) (ans [][]int) {
 	if root == nil {
-		return ans
+		return
 	}
-	var q = []*TreeNode{root}
+	q := []*TreeNode{root}
 	for len(q) > 0 {
-		var t []int
+		t := []int{}
 		for n := len(q); n > 0; n-- {
 			node := q[0]
 			q = q[1:]
@@ -193,7 +201,7 @@ func levelOrder(root *TreeNode) [][]int {
 		}
 		ans = append(ans, t)
 	}
-	return ans
+	return
 }
 ```
 

solution/0100-0199/0102.Binary Tree Level Order Traversal/README_EN.md

@@ -52,10 +52,10 @@ BFS.
 #         self.left = left
 #         self.right = right
 class Solution:
-    def levelOrder(self, root: TreeNode) -> List[List[int]]:
-        if root is None:
-            return []
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
         ans = []
+        if root is None:
+            return ans
         q = deque([root])
         while q:
             t = []
@@ -162,14 +162,13 @@ public:
  *     Right *TreeNode
  * }
  */
-func levelOrder(root *TreeNode) [][]int {
-	var ans [][]int
+func levelOrder(root *TreeNode) (ans [][]int) {
 	if root == nil {
-		return ans
+		return
 	}
-	var q = []*TreeNode{root}
+	q := []*TreeNode{root}
 	for len(q) > 0 {
-		var t []int
+		t := []int{}
 		for n := len(q); n > 0; n-- {
 			node := q[0]
 			q = q[1:]
@@ -183,7 +182,7 @@ func levelOrder(root *TreeNode) [][]int {
 		}
 		ans = append(ans, t)
 	}
-	return ans
+	return
 }
 ```
 

solution/0100-0199/0102.Binary Tree Level Order Traversal/Solution.go

@@ -6,14 +6,13 @@
  *     Right *TreeNode
  * }
  */
-func levelOrder(root *TreeNode) [][]int {
-	var ans [][]int
+func levelOrder(root *TreeNode) (ans [][]int) {
 	if root == nil {
-		return ans
+		return
 	}
-	var q = []*TreeNode{root}
+	q := []*TreeNode{root}
 	for len(q) > 0 {
-		var t []int
+		t := []int{}
 		for n := len(q); n > 0; n-- {
 			node := q[0]
 			q = q[1:]
@@ -27,5 +26,5 @@ func levelOrder(root *TreeNode) [][]int {
 		}
 		ans = append(ans, t)
 	}
-	return ans
+	return
 }

solution/0100-0199/0102.Binary Tree Level Order Traversal/Solution.py

@@ -5,10 +5,10 @@
 #         self.left = left
 #         self.right = right
 class Solution:
-    def levelOrder(self, root: TreeNode) -> List[List[int]]:
-        if root is None:
-            return []
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
         ans = []
+        if root is None:
+            return ans
         q = deque([root])
         while q:
             t = []