0100_same-tree/README.md

@@ -1,34 +0,0 @@
-Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.
-
-Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.
-
-**Example 1:**
-
-![](https://assets.leetcode.com/uploads/2020/12/20/ex1.jpg)
-
-    Input: p = [1,2,3], q = [1,2,3]
-    Output: true
-    
-
-**Example 2:**
-
-![](https://assets.leetcode.com/uploads/2020/12/20/ex2.jpg)
-
-    Input: p = [1,2], q = [1,null,2]
-    Output: false
-    
-
-**Example 3:**
-
-![](https://assets.leetcode.com/uploads/2020/12/20/ex3.jpg)
-
-    Input: p = [1,2,1], q = [1,1,2]
-    Output: false
-    
-
-**Constraints:**
-
-*   The number of nodes in both trees is in the range `[0, 100]`.
-*   `-104 <= Node.val <= 104`
-
-https://leetcode.com/problems/same-tree/

0100_same-tree/python3/solution.py

@@ -1,14 +0,0 @@
-# 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 = right
-class Solution:
-    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
-        if p is None and q is None:
-            return True
-        elif p is None or q is None or p.val != q.val:
-            return False
-        
-        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)

0104_maximum-depth-of-binary-tree/README.md

@@ -1,24 +0,0 @@
-Given the `root` of a binary tree, return _its maximum depth_.
-
-A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.
-
-**Example 1:**
-
-![](https://assets.leetcode.com/uploads/2020/11/26/tmp-tree.jpg)
-
-    Input: root = [3,9,20,null,null,15,7]
-    Output: 3
-    
-
-**Example 2:**
-
-    Input: root = [1,null,2]
-    Output: 2
-    
-
-**Constraints:**
-
-*   The number of nodes in the tree is in the range `[0, 104]`.
-*   `-100 <= Node.val <= 100`
-
-https://leetcode.com/problems/maximum-depth-of-binary-tree

0104_maximum-depth-of-binary-tree/python3/dfs-pre-order-global.py

@@ -1,27 +0,0 @@
-# Time: O(N)
-# Space: O(N)
-
-# 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 = right
-class Solution:
-    maxDepthValue = 0
-    
-    def maxDepth(self, root: Optional[TreeNode]) -> int:
-        '''
-        Recursive DFS pre-order but storing max globally
-        '''
-        def findMax(node, depth):
-            if node is None: return None
-            
-            currDepth = depth + 1
-            self.maxDepthValue = max(currDepth, self.maxDepthValue)
-            findMax(node.left, currDepth)
-            findMax(node.right, currDepth)
-        
-        findMax(root, 0)
-        
-        return self.maxDepthValue

0104_maximum-depth-of-binary-tree/python3/dfs-pre-order-oneliner.py

@@ -1,17 +0,0 @@
-# Time: O(N)
-# Space: O(N)
-
-# 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 = right
-class Solution:
-    maxDepthValue = 0
-    
-    def maxDepth(self, root: Optional[TreeNode]) -> int:
-        '''
-        Recursive DFS
-        '''
-        return 0 if root is None else 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))

0543_diameter-of-binary-tree/README.md

@@ -1,27 +0,0 @@
-Given the `root` of a binary tree, return _the length of the **diameter** of the tree_.
-
-The **diameter** of a binary tree is the **length** of the longest path between any two nodes in a tree. This path may or may not pass through the `root`.
-
-The **length** of a path between two nodes is represented by the number of edges between them.
-
-**Example 1:**
-
-![](https://assets.leetcode.com/uploads/2021/03/06/diamtree.jpg)
-
-    Input: root = [1,2,3,4,5]
-    Output: 3
-    Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].
-    
-
-**Example 2:**
-
-    Input: root = [1,2]
-    Output: 1
-    
-
-**Constraints:**
-
-*   The number of nodes in the tree is in the range `[1, 104]`.
-*   `-100 <= Node.val <= 100`
-
-https://leetcode.com/problems/diameter-of-binary-tree/

0543_diameter-of-binary-tree/python3/solution.py

@@ -1,36 +0,0 @@
-# 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 = right
-class Solution:
-    def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
-        result = [0]
-        
-        def dfs(node):
-            # Leaf node height will be 1
-            # Empty node height will be -1
-            
-            if node is None: return -1
-            
-            lefth = dfs(node.left)
-            righth = dfs(node.right)
-            
-            # Diameter calc:
-            #
-            # Cause for current node will be pointing to both
-            # left and right nodes and we need to consider
-            # them (+2)
-            #
-            # e.g. consider this is leaf node, lefth and righth = -1
-            # so, current node's diameter should be 0 = 2 + -1 + -1
-            result[0] = max(result[0], 2 + lefth + righth)
-            
-            # return max height including the node itself
-            return 1 + max(lefth, righth)
-        
-        dfs(root)
-        
-        return result[0]
-