Create subtree-of-another-tree.py

Author: kamyu104

Python/subtree-of-another-tree.py

@@ -0,0 +1,68 @@
+# Time:  O(m * n), m is the number of nodes of s, n is the number of nodes of t
+# Space: O(h), h is the height of s
+
+# Given two non-empty binary trees s and t,
+# check whether tree t has exactly the same structure and
+# node values with a subtree of s.
+# A subtree of s is a tree consists of a node in s and all of this node's descendants.
+# The tree s could also be considered as a subtree of itself.
+#
+# Example 1:
+# Given tree s:
+#
+#      3
+#     / \
+#   4   5
+#   / \
+#  1   2
+# Given tree t:
+#   4 
+#   / \
+#  1   2
+# Return true, because t has the same structure and node values with a subtree of s.
+# Example 2:
+# Given tree s:
+#
+#      3
+#     / \
+#   4   5
+#   / \
+#  1   2
+#     /
+#   0
+# Given tree t:
+#   4
+#   / \
+#  1   2
+# Return false.
+
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution(object):
+    def isSubtree(self, s, t):
+        """
+        :type s: TreeNode
+        :type t: TreeNode
+        :rtype: bool
+        """
+        def isSame(x, y):
+            if not x and not y:
+                return True
+            if not x or not y:
+                return False
+            return x.val == y.val and \
+                   isSame(x.left, y.left) and \
+                   isSame(x.right, y.right)
+
+        def preOrderTraverse(s, t):
+            return s != None and \
+                   (isSame(s, t) or \
+                    preOrderTraverse(s.left, t) or \
+                    preOrderTraverse(s.right, t))
+        
+        return preOrderTraverse(s, t)