solve: binaryTreeMaximumPathSum, graphValidTree, maximumDepthOfBinaryTree, mergeIntervals and reorderList

Author: yolophg

binary-tree-maximum-path-sum/yolophg.py

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+# Time Complexity: O(N) - visit each node once.
+# Space Complexity: O(H) - recursive call stack, where H is the tree height.
+#                        - O(log N) for balanced trees, O(N) for skewed trees.
+
+class Solution:
+    def maxPathSum(self, root: Optional[TreeNode]) -> int:
+        def dfs(node):
+            nonlocal ans
+            if not node:
+                return 0 
+            
+            # get the max path sum from left and right subtrees (ignore negatives)
+            left = max(dfs(node.left), 0)
+            right = max(dfs(node.right), 0)
+            
+            # update the global max path sum including this node
+            ans = max(ans, left + right + node.val)
+            
+            # return the max path sum that can be extended to the parent
+            return max(left, right) + node.val
+        
+         # initialize with the smallest possible value
+        ans = float('-inf') 
+        # start DFS from the root
+        dfs(root)  
+        return ans

graph-valid-tree/yolophg.py

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+# Time Complexity: O(N) - iterate through all edges and nodes at most once.
+# Space Complexity: O(N) - store the graph as an adjacency list and track visited nodes.
+
+class Solution:    
+    def valid_tree(self, n: int, edges: List[List[int]]) -> bool:
+        # a tree with 'n' nodes must have exactly 'n-1' edges
+        if len(edges) != n - 1:
+            return False
+
+        # build an adjacency list (graph representation)
+        graph = {i: [] for i in range(n)}
+        for u, v in edges:
+            graph[u].append(v)
+            graph[v].append(u)
+
+        # use BFS to check if the graph is fully connected and acyclic
+        visited = set()
+        queue = [0]
+        visited.add(0)
+
+        while queue:
+            node = queue.pop(0)  
+            for neighbor in graph[node]:
+                if neighbor in visited:
+                    continue
+                visited.add(neighbor)
+                queue.append(neighbor)
+
+        # if we visited all nodes, it's a valid tree
+        return len(visited) == n

maximum-depth-of-binary-tree/yolophg.py

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+# Time Complexity: O(N) - visit each node once.
+# Space Complexity: O(H) - the recursion stack goes as deep as the height of the tree.
+#                          - Worst case (skewed tree): O(N)
+#                          - Best case (balanced tree): O(log N)
+
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:        
+        # if there's no node, the depth is just 0.
+        if not root:
+            return 0
+        
+        # recursively get the depth of left and right subtrees
+        # then add 1 for the current node
+        return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))

merge-intervals/yolophg.py

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+# Time Complexity: O(N log N) - sorting takes O(N log N), and merging takes O(N)
+# Space Complexity: O(N) - store the merged intervals in a new list.
+    
+class Solution:
+
+    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
+        # sort the intervals by their start values
+        intervals.sort(key=lambda x: x[0])
+
+        # this will store our final merged intervals
+        merged = [] 
+        # start with the first interval
+        prev = intervals[0]  
+        
+        # iterate through the sorted intervals
+        for interval in intervals[1:]:
+            if interval[0] <= prev[1]:  
+                # overlapping intervals → merge them by updating 'end' value
+                prev[1] = max(prev[1], interval[1])
+            else:
+                # no overlap → add 'prev' to the merged list and update 'prev'
+                merged.append(prev)
+                prev = interval
+
+        merged.append(prev)
+        
+        return merged

reorder-list/yolophg.py

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+# Time Complexity: O(N) - traverse the list three times: 
+#                       (1) find the middle (O(N)), 
+#                       (2) reverse the second half (O(N)), 
+#                       (3) merge the two halves (O(N)).
+# Space Complexity: O(1) - use a few extra pointers, no additional data structures.
+
+class Solution:
+    def reorderList(self, head: Optional[ListNode]) -> None:
+        # find the middle of the list using slow & fast pointers
+        slow, fast = head, head.next
+        while fast and fast.next:
+            slow = slow.next
+            fast = fast.next.next
+
+        # reverse the second half of the list
+        second = slow.next
+        # cut the list into two halves
+        slow.next = None  
+        prev = None
+        
+        while second:
+            temp = second.next
+            second.next = prev
+            prev = second
+            second = temp
+        
+        # merge the two halves (alternating nodes)
+        first, second = head, prev
+
+        while second:
+            temp1, temp2 = first.next, second.next
+            first.next = second
+            second.next = temp1
+            first, second = temp1, temp2