Completed Trees-3

Problem1.py

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+# Problem1 (https://leetcode.com/problems/path-sum-ii/)
+# Time Complexity: O(n), we visit every node once; building each saved path costs O(h) and there can be up to O(n) such paths in the worst case (skewed tree)
+# Space Complexity: O(h), h is the tree height; that's how deep the recursion stack and the shared "path" list go (output space for result not counted)
+
+# Approach:
+# We walk down the tree from root to leaf using recursion, keeping ONE shared list called "path" that holds the values of the current branch we're standing on.
+# At every node we do 3 things: add it to the running sum add it to "path"and check if we've hit a leaf with the matching sum (if so, save a COPY of the path).
+# Then we go explore the left child completely, then the right child completely.
+# Once both children are fully done, we "pop" (remove) the current node from "path" before going back to the parent (this is called backtracking.)
+# It undoes our own change so the path is clean and correct for whichever branch gets explored next.
+
+# 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 pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
+        self.result = []                          # empty box to collect all winning paths
+        self.helper(root, targetSum, 0, [])       # start walk: sum so far = 0, path so far = empty
+        return self.result                        # give back everything we found
+
+    def helper(self, root, targetSum, currSum, path):
+        if root is None:                          # no node here (empty spot, ran off the tree)
+            return                                # nothing to do, go back to caller
+
+        currSum = root.val + currSum              # add this node's value to the running sum
+        path.append(root.val)                     # add this node's value to the shared path
+
+        if root.left is None and root.right is None:   # this node has no children, it's a leaf
+            if currSum == targetSum:                    # check if the sum matches target
+                self.result.append(list(path))           # save a COPY of path (not the same list!)
+
+        self.helper(root.left, targetSum, currSum, path)    # fully explore the left side first
+        self.helper(root.right, targetSum, currSum, path)   # then fully explore the right side
+        path.pop()                                # done with this node, remove it before going back(undoes the append from above — keeps path clean for the next branch the parent explores)

Problem2.py

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+# Problem2 (https://leetcode.com/problems/symmetric-tree/)
+# Time Complexity: O(n),we visit every node in the tree exactly once
+# Space Complexity: O(h),recursion stack goes as deep as the tree's height h
+
+# Logic:
+# 1. A tree is symmetric if its left subtree mirrors its right subtree.
+# 2. Two nodes mirror each other when their values match AND their outer/inner children mirror.
+# 3. We recurse with cross-pairing (left.left with right.right) to compare the mirror directly.
+
+# 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 isSymmetric(self, root: Optional[TreeNode]) -> bool:
+        if root is None:                                 # empty tree is symmetric by default
+            return True                                  # nothing to compare, so return True
+        return self.helper(root.left, root.right)        # ask: do root's two children mirror each other?
+
+    def helper(self, left, right):
+        if left is None and right is None:               # both sides empty at the same spot
+            return True                                  # they match perfectly, this pair is fine
+        if left is None or right is None:                # exactly one side is empty (other isn't)
+            return False                                 # one node has nothing to mirror,not symmetric
+        if left.val != right.val:                        # both nodes exist but values differ
+            return False                                 # mismatch found, not symmetric
+
+        # cross-pair the children: outer with outer, inner with inner
+        return self.helper(left.left, right.right) and self.helper(left.right, right.left)
+        #                       outer pair                             inner pair 
+        #both pairs must mirror (and) for this pair to be symmetric