no message

Author: Chris Wu

problems/diameter-of-binary-tree.py

@@ -45,6 +45,9 @@ def getMaxDepth(node):
 From an unknown node, that its max_depth_from_left (`l`) + max_depth_from_right (`r`) is the biggest.
 The node that generate this number could be from any node, so we iterate through every node to update `ans`.
 In other words, to find the answer, we need to check every node, if the max diameter pass through here.
+
+Time complexity is O(N), where N is the number of nodes.
+Space complexity is O(LogN), since we might got to LogN level on recursion.
 """
 
 # from collections import defaultdict

problems/sum-root-to-leaf-numbers.py

@@ -8,11 +8,16 @@ def sumNumbers(self, root):
         while stack:
             node, total = stack.pop()
             total += str(node.val)
-            if not node.left and not node.right:
+            if not node.left and not node.right: #is_leaf
                 ans += int(total)
                 continue
             if node.left:
                 stack.append((node.left, total))
             if node.right:
                 stack.append((node.right, total))
         return ans
+
+"""
+Time complexity: O(N), since we use DFS to traverse each node once.
+Space complexity: O(NLogN), since each node (at the very bottom), may carry all the digit from its ancestor, LogN.
+"""