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.
Solution:
1.method use recursive and compare whether it is the same tree for each substree rooted at s' and the tree t
it takes o(mn) time ; m is the size of s, n is the size of t
2. use preorder traversal and innorder traversal; o(m+n)
Here I analysis why for both traversal.
for inorder traversal
for example 1: s list [ #, 1, #, 4, #, 2, #,3, #, 5, #] , t list [#, 1, #, 4, #, 2, #]
example 2: s list [#, 1,#, 4, #, 0, #, 2, #, 3, 5, #, #], t list [ #, 1, #, 4, #, 2, #]
example 3: s list [ #, 4, #, 3, #, 5, #, 1, #, 2, #]; [#, 4, #, 3, #]; it is the sublist of s, but not the subtree of s, does not work for inorder
but for the preorder of this, s list [1, 4, #, 3, #, 5, #, #, 2, #, #], t list [4, #,3, #, #] ; it works.
1 def helperRecursiveInOrder(root, l): 2 if root is None: 3 l.append (str("#")) 4 else: 5 helperRecursiveInOrder(root.left, l) 6 l.append(str(root.val)) 7 helperRecursiveInOrder(root.right, l) 8 9 10 def helperRecursivePreOrder(root, l): 11 if root is None: 12 l.append (str("#")) 13 else: 14 l.append(str(root.val)) 15 helperRecursivePreOrder(root.left, l) 16 helperRecursivePreOrder(root.right, l) 17 18 19 lsIn = [] 20 helperRecursiveInOrder(s, lsIn) 21 ltIn = [] 22 helperRecursiveInOrder(t, ltIn) 23 strSIn = ",".join(lsIn) 24 strTIn = ",".join(ltIn) 25 26 lsPre = [] 27 helperRecursivePreOrder(s, lsPre) 28 ltPre = [] 29 helperRecursivePreOrder(t, ltPre) 30 strSPre = ",".join(lsPre) 31 strTPre = ",".join(ltPre) 32 #print ("str: ", strSPre, strTPre, lsPre) 33 if(strTIn in strSIn and strTPre in strSPre): 34 return True 35 else: 36 return False
3. After finishing this, I checked the answer online, actually, it used preorder traversal only to tackle this problem.
if we simply process the case like s = [12] , t =[2]; with the ',' added to each node,
with the preOrder, it also works.
Refer to the work from http://www.cnblogs.com/grandyang/p/6828687.html
1 def helperRecursivePreOrder(root, l): 2 if root is None: 3 l.append (str("#")) 4 else: 5 l.append(',' + str(root.val)) 6 helperRecursivePreOrder(root.left, l) 7 helperRecursivePreOrder(root.right, l) 8 lsPre = [] 9 helperRecursivePreOrder(s, lsPre) 10 ltPre = [] 11 helperRecursivePreOrder(t, ltPre) 12 strSPre = ",".join(lsPre) 13 strTPre = ",".join(ltPre) 14 #print ("str: ", strSPre, strTPre, lsPre) 15 if(strTPre in strSPre): 16 return True 17 else: 18 return False 19