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Leetcode binary tree inorder traversal #10
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42fb596
+linked-list.md +reverse-linked-list
Igneaalis 080e833
Changed variables' name, added hyperlinks (anchors)
Igneaalis 0d5a662
Improved [Reverse linked list]
Igneaalis 57b1910
Added LEETCODE middle-of-the-linked-list solution.
Igneaalis 303c7ae
Added LEETCODE palindrome-linked-list solution.
Igneaalis cbc585f
Added LEETCODE merge-two-sorted-lists solution.
Igneaalis 07c4497
Added LEETCODE Merge Two Sorted Lists (iterative) solution.
Igneaalis 5d394cc
Added a forgotten link for an iterative solution.
Igneaalis f6c5605
Added LEETCODE "binary-tree-inorder-traversal" solution.
Igneaalis 4101f7e
Added recursive and iterative,stack LEETCODE "binary-tree-inorder-tra…
Igneaalis 164fd7e
Removed an extra line from recursive solution.
Igneaalis bee6f14
Removed an extra line from recursive solution (another).
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,98 @@ | ||
| # Binary tree | ||
|
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| + [Binary Tree Inorder Traversal (iterative, threaded)](#binary-tree-inorder-traversal-iterative-threaded) | ||
| + [Binary Tree Inorder Traversal (recursive)](#binary-tree-inorder-traversal-recursive) | ||
| + [Binary Tree Inorder Traversal (iterative, stack)](#binary-tree-inorder-traversal-iterative-stack) | ||
|
|
||
| ## Binary Tree Inorder Traversal (iterative, threaded) | ||
|
|
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| https://leetcode.com/problems/binary-tree-inorder-traversal/ | ||
|
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||
| ```python | ||
| # Definition for a binary tree node. | ||
| # class TreeNode(object): | ||
| # def __init__(self, val=0, left=None, right=None): | ||
| # self.val = val | ||
| # self.left = left | ||
| # self.right = right | ||
| class Solution(object): | ||
| def inorderTraversal(self, root): | ||
| """ | ||
| :type root: TreeNode | ||
| :rtype: List[int] | ||
| """ | ||
| head = root | ||
| inorderlist = [] | ||
| while head: | ||
| if not head.left: | ||
| inorderlist.append(head.val) | ||
| head = head.right | ||
| elif head.left: | ||
| curNode = head.left | ||
| while head.left and curNode.right and curNode.right != head: | ||
| curNode = curNode.right | ||
| if not curNode.right: | ||
| curNode.right = head | ||
| head = head.left | ||
| elif curNode.right == head: | ||
| inorderlist.append(head.val) | ||
| curNode.right = None | ||
| head = head.right | ||
| return inorderlist | ||
|
|
||
| ``` | ||
|
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||
| ## Binary Tree Inorder Traversal (recursive) | ||
|
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||
| https://leetcode.com/problems/binary-tree-inorder-traversal/ | ||
|
|
||
| ```python | ||
| # Definition for a binary tree node. | ||
| # class TreeNode(object): | ||
| # def __init__(self, val=0, left=None, right=None): | ||
| # self.val = val | ||
| # self.left = left | ||
| # self.right = right | ||
| class Solution(object): | ||
| def inorderTraversal(self, root): | ||
| """ | ||
| :type root: TreeNode | ||
| :rtype: List[int] | ||
| """ | ||
| head = root | ||
| return self.inorderTraversal(head.left) + [head.val] + self.inorderTraversal(head.right) if head else [] | ||
|
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| ``` | ||
|
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| ## Binary Tree Inorder Traversal (iterative, stack) | ||
|
|
||
| https://leetcode.com/problems/binary-tree-inorder-traversal/ | ||
|
|
||
| ```python | ||
| # Definition for a binary tree node. | ||
| # class TreeNode(object): | ||
| # def __init__(self, val=0, left=None, right=None): | ||
| # self.val = val | ||
| # self.left = left | ||
| # self.right = right | ||
| class Solution(object): | ||
| def inorderTraversal(self, root): | ||
| """ | ||
| :type root: TreeNode | ||
| :rtype: List[int] | ||
| """ | ||
| if not root: | ||
| return [] | ||
| stack = [] | ||
| inorderlist = [] | ||
| while stack or root: | ||
| if root: | ||
| stack.append(root) | ||
| root = root.left | ||
| else: | ||
| node = stack.pop() | ||
| inorderlist.append(node.val) | ||
| root = node.right | ||
| return inorderlist | ||
|
|
||
| ``` | ||
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Очень круто, но много кода. Ты такой код не успеешь написать быстро, много разных if. Можешь оставить
Но напиши рядом рекурсивное решение и решение, где ты рекурсию раскроешь через stack, т.е. получится итеративное решение без лишних if
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Наличие многих if - место для потенциальной ошибки. Решения здесь проще
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@Vivelapaix добавил два решения. Ещё вопрос по гиту. Как изменить имя запушенного коммита?