Update 0102.二叉树的层序遍历.md

problems/0102.二叉树的层序遍历.md

@@ -98,7 +98,7 @@ class Solution:
         out_list = []
 
         while quene:
-            length = len(queue)  # 这里一定要先求出队列的长度，不能用range(len(queue))，因为queue长度是变化的
+            length = len(queue)  
             in_list = []
             for _ in range(length):
                 curnode = queue.pop(0)  # （默认移除列表最后一个元素）这里需要移除队列最头上的那个
@@ -627,6 +627,27 @@ public:
     }
 };
 ```
+python代码：
+
+```python
+class Solution:
+    def largestValues(self, root: TreeNode) -> List[int]:
+        if root is None:
+            return []
+        queue = [root]
+        out_list = []
+        while queue:
+            length = len(queue)
+            in_list = []
+            for _ in range(length):
+                curnode = queue.pop(0)
+                in_list.append(curnode.val)
+                if curnode.left: queue.append(curnode.left)
+                if curnode.right: queue.append(curnode.right)
+            out_list.append(max(in_list))
+        return out_list
+```
+
 javascript代码：
 
 ```javascript
@@ -712,6 +733,42 @@ public:
 };
 ```
 
+python代码：
+
+```python
+# 层序遍历解法
+class Solution:
+    def connect(self, root: 'Node') -> 'Node':
+        if not root:
+            return None
+        queue = [root]
+        while queue:
+            n = len(queue)
+            for i in range(n):
+                node = queue.pop(0)
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+                if i == n - 1:
+                    break
+                node.next = queue[0]
+        return root
+
+# 链表解法
+class Solution:
+    def connect(self, root: 'Node') -> 'Node':
+        first = root
+        while first:
+            cur = first
+            while cur:  # 遍历每一层的节点
+                if cur.left: cur.left.next = cur.right  # 找左节点的next
+                if cur.right and cur.next: cur.right.next = cur.next.left  # 找右节点的next
+                cur = cur.next # cur同层移动到下一节点
+            first = first.left  # 从本层扩展到下一层
+        return root
+```
+
 ## 117.填充每个节点的下一个右侧节点指针II
 
 题目地址：https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node-ii/
@@ -753,7 +810,48 @@ public:
     }
 };
 ```
+python代码：
 
+```python
+# 层序遍历解法
+class Solution:
+    def connect(self, root: 'Node') -> 'Node':
+        if not root:
+            return None
+        queue = [root]
+        while queue:  # 遍历每一层
+            length = len(queue)
+            tail = None # 每一层维护一个尾节点
+            for i in range(length): # 遍历当前层
+                curnode = queue.pop(0)
+                if tail:
+                    tail.next = curnode # 让尾节点指向当前节点
+                tail = curnode # 让当前节点成为尾节点
+                if curnode.left : queue.append(curnode.left)
+                if curnode.right: queue.append(curnode.right)
+        return root
+
+# 链表解法
+class Solution:
+    def connect(self, root: 'Node') -> 'Node':
+        if not root:
+            return None
+        first = root
+        while first:  # 遍历每一层
+            dummyHead = Node(None)  # 为下一行创建一个虚拟头节点，相当于下一行所有节点链表的头结点(每一层都会创建)；
+            tail = dummyHead  # 为下一行维护一个尾节点指针（初始化是虚拟节点）
+            cur = first
+            while cur:  # 遍历当前层的节点
+                if cur.left:  # 链接下一行的节点
+                    tail.next = cur.left
+                    tail = tail.next
+                if cur.right:
+                    tail.next = cur.right
+                    tail = tail.next
+                cur = cur.next  # cur同层移动到下一节点
+            first = dummyHead.next  # 此处为换行操作，更新到下一行
+        return root
+```
 
 ## 总结