added 0100_same_tree.py

Author: nyierr

problems/easy/0100_same_tree.py

@@ -0,0 +1,78 @@
+"""
+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:
+  Input: p = [1,2,3], q = [1,2,3]
+  Output: true
+    
+Example 2:
+  Input: p = [1,2], q = [1,null,2]
+  Output: false
+
+Example 3:
+  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]
+  * -10^4 <= Node.val <= 10^4
+"""
+
+class Solution:          
+    # O(2^n) solution, iterative
+    #def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+    #    if p is None and q is None:
+    #        return True
+    #    
+    #    if p is None or q is None:
+    #        return False
+    #    
+    #    stack_p = [p.right, p.left, p]
+    #    stack_q = [q.right, q.left, q]
+    #    while len(stack_p) and len(stack_q):
+    #        p = stack_p.pop()
+    #        q = stack_q.pop()
+    #    
+    #        if p is None and q is None:
+    #            return True
+    #    
+    #        if p.val != q.val:
+    #            return False
+    #
+    #        # False, if one has a left node and the other one does not
+    #        if (p.left is None) ^ (q.left is None):
+    #            return False
+    #        
+    #        # False, if one has a right node and the other one does not
+    #        if (p.right is None) ^ (q.right is None):
+    #            return False
+    #        
+    #        if not p.right is None:
+    #            stack_p.append(p.right)
+    #            stack_q.append(q.right)
+    #            
+    #        if not p.left is None:
+    #            stack_p.append(p.left)
+    #            stack_q.append(q.left)
+    #        
+    #    return len(stack_p) == len(stack_q)
+    
+    # O(2^n) solution, recursive one
+    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+        if (p is None) ^ (q is None):
+            return False
+            
+        if p is None and q is None:
+            return True
+        
+        if p.val != q.val:
+            return False
+        
+        if self.isSameTree(p.left, q.left):
+            return self.isSameTree(p.right, q.right)
+            
+        return False