O(n) time and O(n) space to hold n elements in a deque.

149ps · Dec 25, 2019 · 995288a · 995288a
1 parent 8c8446c
commit 995288a
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Showing 2 changed files with 38 additions and 22 deletions.
diff --git a/102_Binary_Tree_Level_Order_Traversal.py → .../102_Binary_Tree_Level_Order_Traversal.py b/102_Binary_Tree_Level_Order_Traversal.py → .../102_Binary_Tree_Level_Order_Traversal.py
@@ -10,22 +10,22 @@ def levelOrder(self, root: TreeNode) -> List[List[int]]:
         if not root:
             return None
         q = collections.deque()
+        node = None
         q.append(root)
         q.append(None)
-        node = None
         res = []
         temp = []
         while q:
             node = q.popleft()
-            if node == None:
-                res.append(temp)
-                temp = []
-                if q:
-                    q.append(None)
-            else:
+            if node:
                 temp.append(node.val)
                 if node.left:
                     q.append(node.left)
-                if node.right != None:
+                if node.right:
                     q.append(node.right)
-        return res
+            else:
+                res.append(temp)
+                temp = []
+                if q:       # if there are still elements in a queue
+                    q.append(None)
+        return res
diff --git a/107_Binary_Tree_Level_Order_Traversal_II.py → ...7_Binary_Tree_Level_Order_Traversal_II.py b/107_Binary_Tree_Level_Order_Traversal_II.py → ...7_Binary_Tree_Level_Order_Traversal_II.py
@@ -1,3 +1,20 @@
+"""
+Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
+
+For example:
+Given binary tree [3,9,20,null,null,15,7],
+    3
+   / \
+  9  20
+    /  \
+   15   7
+return its bottom-up level order traversal as:
+[
+  [15,7],
+  [9,20],
+  [3]
+]
+"""
 # Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):
@@ -7,29 +24,28 @@
 
 class Solution:
     def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
+        q = collections.deque()
+        node = None
         if not root:
             return None
-        p = collections.deque()
         q = collections.deque()
+        node = None
         q.append(root)
         q.append(None)
-        node = None
         res = []
         temp = []
         while q:
             node = q.popleft()
-            if node == None:
-                p.append(temp)
-                temp = []
-                if q:
-                    q.append(None)
-            else:
+            if node:
                 temp.append(node.val)
                 if node.left:
                     q.append(node.left)
-                if node.right != None:
+                if node.right:
                     q.append(node.right)
-        while p:
-            res.append(p.pop())
-        return res
-
+            else:
+                res.append(temp)
+                temp = []
+                if q:       # if there are still elements in a queue
+                    q.append(None)
+        return reversed(res)
+