572. Subtree of Another Tree

https://leetcode.com/problems/subtree-of-another-tree/#/solutions

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.

Sol:

# 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 check(s, t):
            # helper function that does the actual subtree check
            if (s is None) and (t is None):
                return True
            if (s is None) or (t is None):
                return False
            return (s.val == t.val and check(s.left, t.left) and check(s.right, t.right))

        # need to do a pre-order traversal and do a check
        # for every node we visit for the subtree
        if not s:
            # return False since None cannot contain a subtree 
            return
        if check(s, t):
            # we found a match
            return True
        if self.isSubtree(s.left, t) or self.isSubtree(s.right, t):
            # a match was found
            return True
        return False
原文地址:https://www.cnblogs.com/prmlab/p/6961620.html