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목록tree (14)
Tech for good
[Leetcode/Tree, Binary Tree, DFS] 572. Subtree of Another Tree
# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool: if not root: return False return self.sameTree(root, subRoot) or self.isSubtree(root..
[Leetcode/Tree, Binary Search Tree, DFS] 1305. All Elements in Two Binary Search Trees
# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def getAllElements(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> List[int]: def helper(root): if not root: return [] return helper(r..
[Leetcode/Tree, Binary Tree, DFS] 110. Balanced Binary Tree
# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def isBalanced(self, root: Optional[TreeNode]) -> bool: def check(node): if not node: return 0, True # height = 0, isBalanced = True lef..
[Leetcode/Tree, DFS, Binary Tree] 543. Diameter of Binary Tree
# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int: def helper(node): if not node: return 0, 0 left_h, left_d = helper(node...
[Leetcode/Tree, Binary Tree, Stack, DFS] 94. Binary Tree Inorder Traversal
Tree > Traversal 1. DFS => Recursion (using self; recursive) or Stack (iterative)Pre-order: Root -> Left -> RightIn-order: Left -> Root -> RightPost-order: Left -> Right -> Root# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def inorderTrave..
[Leetcode/Tree, Binary Tree, Binary Search Tree, Recursion] 108. Convert Sorted Array to Binary Search Tree
Solution1: using pointers (w/ helper function)# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def sortedArrayToBST(self, nums: List[int]) -> TreeNode: def helper(left, right): if left > right: return None ..
[Leetcode/Tree, DFS, Binary Tree, Recursion] 112. Path Sum
Binary TreeBinary Search Tree AscendingLeft to Right # Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool: if root is None: return False if not root.l..
[Leetcode/Tree, DFS, Binary Tree, Recursion] 226. Invert Binary Tree
# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]: # Base case: decide where to stop if root is None: return None right = self.invertT..
[Leetcode/Stack,Tree] 897. Increasing Order Search Tree
✅ Iterative in-order traversal using a stackclass Solution: def increasingBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]: stack = [] dummy = TreeNode(-1) # Dummy node to build the result tree curr = dummy # Pointer to build the new tree node = root # Start traversal from the root while stack or node: # 1. Go ..
[Leetcode/Tree] 617. Merge Two Binary Trees
# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]: if not root1: # root1이 None이면 root2 반환 return root2 if not root2: # ro..