From 186d4e4ca41d2de86e25a009bb89d454096e5405 Mon Sep 17 00:00:00 2001 From: Wen Liang Date: Thu, 2 Sep 2021 14:06:04 +0800 Subject: [PATCH 1/5] =?UTF-8?q?=E4=BF=AE=E6=AD=A3=20=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E8=BF=AD=E4=BB=A3=E9=81=8D=E5=8E=86.md=20?= =?UTF-8?q?=E4=B8=AD=E7=9A=84=E6=8B=BC=E5=86=99=E9=94=99=E8=AF=AF=20?= =?UTF-8?q?=E8=B7=9F=E8=8A=82=E7=82=B9=20->=20=E6=A0=B9=E8=8A=82=E7=82=B9?= =?UTF-8?q?=20=E4=B8=8D=E6=83=B3=20=20=20->=20=E4=B8=8D=E5=83=8F?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树的迭代遍历.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/二叉树的迭代遍历.md b/problems/二叉树的迭代遍历.md index a1d65070..84363610 100644 --- a/problems/二叉树的迭代遍历.md +++ b/problems/二叉树的迭代遍历.md @@ -27,7 +27,7 @@ 我们先看一下前序遍历。 -前序遍历是中左右，每次先处理的是中间节点，那么先将跟节点放入栈中，然后将右孩子加入栈，再加入左孩子。 +前序遍历是中左右，每次先处理的是中间节点，那么先将根节点放入栈中，然后将右孩子加入栈，再加入左孩子。 为什么要先加入 右孩子，再加入左孩子呢？ 因为这样出栈的时候才是中左右的顺序。 @@ -140,7 +140,7 @@ public: # 总结 -此时我们用迭代法写出了二叉树的前后中序遍历，大家可以看出前序和中序是完全两种代码风格，并不想递归写法那样代码稍做调整，就可以实现前后中序。 +此时我们用迭代法写出了二叉树的前后中序遍历，大家可以看出前序和中序是完全两种代码风格，并不像递归写法那样代码稍做调整，就可以实现前后中序。 **这是因为前序遍历中访问节点（遍历节点）和处理节点（将元素放进result数组中）可以同步处理，但是中序就无法做到同步！** From f5a5cd882f3725cc56a95541bb738025e7b05d89 Mon Sep 17 00:00:00 2001 From: Wen Liang Date: Thu, 2 Sep 2021 17:22:28 +0800 Subject: [PATCH 2/5] =?UTF-8?q?=E4=BC=98=E5=8C=96=200102.=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?.md=20Python3=E8=A7=A3=E6=B3=95=20=E4=BB=BB=E4=BD=95=E5=9C=A8?= =?UTF-8?q?=20list=20=E5=A4=B4=E9=83=A8=E8=BF=9B=E8=A1=8C=E7=9A=84?= =?UTF-8?q?=E6=93=8D=E4=BD=9C=E9=83=BD=E4=BC=9A=E6=8D=9F=E5=A4=B1=E4=B8=80?= =?UTF-8?q?=E5=AE=9A=E6=80=A7=E8=83=BD=20=E5=9C=A8=20Python=20=E4=B8=AD?= =?UTF-8?q?=E5=BA=94=E8=AF=A5=E4=BD=BF=E7=94=A8=20collections.deque=20?= =?UTF-8?q?=E4=BD=9C=E4=B8=BA=E9=98=9F=E5=88=97=E7=9A=84=E6=95=B0=E6=8D=AE?= =?UTF-8?q?=E7=B1=BB=E5=9E=8B=20=E5=9C=A8=E8=AF=A5=E6=95=B0=E6=8D=AE?= =?UTF-8?q?=E7=B1=BB=E5=9E=8B=E5=B7=A6=E5=8F=B3=E4=B8=A4=E7=AB=AF=E8=BF=BD?= =?UTF-8?q?=E5=8A=A0=E5=92=8C=E5=BC=B9=E5=87=BA=E5=85=83=E7=B4=A0=E7=9A=84?= =?UTF-8?q?=E6=97=B6=E9=97=B4=E5=A4=8D=E6=9D=82=E5=BA=A6=E9=83=BD=E6=8E=A5?= =?UTF-8?q?=E8=BF=91O(1)?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 37 +++++++++++++---------- 1 file changed, 21 insertions(+), 16 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index a57a92aa..07708e5a 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -87,26 +87,31 @@ public: python代码： -```python +```python3 class Solution: + """二叉树层序遍历迭代解法""" + def levelOrder(self, root: TreeNode) -> List[List[int]]: + results = [] if not root: - return [] + return results + + from collections import deque + que = deque([root]) + + while que: + size = len(que) + result = [] + for _ in range(size): + cur = que.popleft() + result.append(cur.val) + if cur.left: + que.append(cur.left) + if cur.right: + que.append(cur.right) + results.append(result) - queue = [root] - out_list = [] - - while queue: - 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 - in_list.append(curnode.val) - if curnode.left: queue.append(curnode.left) - if curnode.right: queue.append(curnode.right) - out_list.append(in_list) - - return out_list + return results ``` java: From acd13fc7d81aae428ee7bfd735b9e06c593eee11 Mon Sep 17 00:00:00 2001 From: Wen Date: Thu, 2 Sep 2021 20:55:02 +0800 Subject: [PATCH 3/5] =?UTF-8?q?=E4=BC=98=E5=8C=96=200102.=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?.md=20=E4=B8=AD=E7=9A=84=20107.=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E5=B1=82=E6=AC=A1=E9=81=8D=E5=8E=86=20II=20Python3?= =?UTF-8?q?=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 40 +++++++++++------------ 1 file changed, 20 insertions(+), 20 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 07708e5a..a02cf997 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -279,29 +279,29 @@ python代码： ```python class Solution: + """二叉树层序遍历II迭代解法""" + def levelOrderBottom(self, root: TreeNode) -> List[List[int]]: + results = [] if not root: - return [] - quene = [root] - out_list = [] + return results - while quene: - in_list = [] - for _ in range(len(quene)): - node = quene.pop(0) - in_list.append(node.val) - if node.left: - quene.append(node.left) - if node.right: - quene.append(node.right) - - out_list.append(in_list) - - out_list.reverse() - return out_list - -# 执行用时：36 ms, 在所有 Python3 提交中击败了92.00%的用户 -# 内存消耗：15.2 MB, 在所有 Python3 提交中击败了63.76%的用户 + from collections import deque + que = deque([root]) + + while que: + result = [] + for _ in range(len(que)): + cur = que.popleft() + result.append(cur.val) + if cur.left: + que.append(cur.left) + if cur.right: + que.append(cur.right) + results.append(result) + + results.reverse() + return results ``` Java： From fde18e8fddd5e2686ab46157344b5b3bbdb3d715 Mon Sep 17 00:00:00 2001 From: Wen Liang Date: Fri, 3 Sep 2021 08:36:36 +0800 Subject: [PATCH 4/5] =?UTF-8?q?=E4=BC=98=E5=8C=96=200102.=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?.md=20=E4=B8=AD=E7=9A=84=20637.=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E5=B1=82=E5=B9=B3=E5=9D=87=E5=80=BC=20Python3?= =?UTF-8?q?=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 41 +++++++++++------------ 1 file changed, 19 insertions(+), 22 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 07708e5a..4a0f62e0 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -633,32 +633,29 @@ python代码： ```python class Solution: + """二叉树层平均值迭代解法""" + def averageOfLevels(self, root: TreeNode) -> List[float]: + results = [] if not root: - return [] + return results - quene = deque([root]) - out_list = [] - - while quene: - in_list = [] - - for _ in range(len(quene)): - node = quene.popleft() - in_list.append(node.val) - if node.left: - quene.append(node.left) - if node.right: - quene.append(node.right) - - out_list.append(in_list) - - out_list = map(lambda x: sum(x) / len(x), out_list) - - return out_list + from collections import deque + que = deque([root]) -# 执行用时：56 ms, 在所有 Python3 提交中击败了81.48%的用户 -# 内存消耗：17 MB, 在所有 Python3 提交中击败了89.68%的用户 + while que: + size = len(que) + sum_ = 0 + for _ in range(size): + cur = que.popleft() + sum_ += cur.val + if cur.left: + que.append(cur.left) + if cur.right: + que.append(cur.right) + results.append(sum_ / size) + + return results ``` java: From 578e9908c5470016110e8668815871c9826ff248 Mon Sep 17 00:00:00 2001 From: Wen Liang Date: Fri, 3 Sep 2021 09:24:10 +0800 Subject: [PATCH 5/5] =?UTF-8?q?=E4=BC=98=E5=8C=96=200102.=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?.md=20=E4=B8=AD=E7=9A=84=20429.N=E5=8F=89=E6=A0=91=E7=9A=84?= =?UTF-8?q?=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86=20Python3=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 58 +++++++---------------- 1 file changed, 17 insertions(+), 41 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 4a0f62e0..c25dbda4 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -825,52 +825,28 @@ public: python代码： ```python - class Solution: + """N叉树的层序遍历迭代法""" + def levelOrder(self, root: 'Node') -> List[List[int]]: + results = [] if not root: - return [] + return results - quene = deque([root]) - out_list = [] - - while quene: - in_list = [] - - for _ in range(len(quene)): - node = quene.popleft() - in_list.append(node.val) - if node.children: - # 这个地方要用extend而不是append，我们看下面的例子： - # In [18]: alist=[] - # In [19]: alist.append([1,2,3]) - # In [20]: alist - # Out[20]: [[1, 2, 3]] - # In [21]: alist.extend([4,5,6]) - # In [22]: alist - # Out[22]: [[1, 2, 3], 4, 5, 6] - # 可以看到extend对要添加的list进行了一个解包操作 - # print(root.children)，可以得到children是一个包含 - # 孩子节点地址的list，我们使用for遍历quene的时候， - # 希望quene是一个单层list，所以要用extend - # 使用extend的情况，如果print(quene),结果是 - # deque([<__main__.Node object at 0x7f60763ae0a0>]) - # deque([<__main__.Node object at 0x7f607636e6d0>, <__main__.Node object at 0x7f607636e130>, <__main__.Node object at 0x7f607636e310>]) - # deque([<__main__.Node object at 0x7f607636e880>, <__main__.Node object at 0x7f607636ef10>]) - # 可以看到是单层list - # 如果使用append，print(quene)的结果是 - # deque([<__main__.Node object at 0x7f18907530a0>]) - # deque([[<__main__.Node object at 0x7f18907136d0>, <__main__.Node object at 0x7f1890713130>, <__main__.Node object at 0x7f1890713310>]]) - # 可以看到是两层list，这样for的遍历就会报错 - - quene.extend(node.children) - - out_list.append(in_list) + from collections import deque + que = deque([root]) - return out_list - -# 执行用时：60 ms, 在所有 Python3 提交中击败了76.99%的用户 -# 内存消耗：16.5 MB, 在所有 Python3 提交中击败了89.19%的用户 + while que: + result = [] + for _ in range(len(que)): + cur = que.popleft() + result.append(cur.val) + # cur.children 是 Node 对象组成的列表，也可能为 None + if cur.children: + que.extend(cur.children) + results.append(result) + + return results ``` java: