剑指Offer(Python多种思路实现):树的子结构

面试26题：

题目：树的子结构

题：输入两棵二叉树A和B，判断B是不是A的子结构。

解题思路一：递归，注意空指针的情况。

class Solution:
    def HasSubtree(self, pRoot1, pRoot2):
        # write code here
        res=False
        if pRoot1 and pRoot2:
            if pRoot1.val==pRoot2.val:
                res=self.SubtreeCore(pRoot1,pRoot2)
            if not res:
                res=self.HasSubtree(pRoot1.left,pRoot2)
            if not res:
                res=self.HasSubtree(pRoot1.right,pRoot2)
        return res
    
    def SubtreeCore(self,pRoot1,pRoot2):
        if pRoot2==None:
            return True
        if pRoot1==None:
            return False
        if pRoot1.val!=pRoot2.val:
            return False
        return self.SubtreeCore(pRoot1.left,pRoot2.left) and self.SubtreeCore(pRoot1.right,pRoot2.right)

解题思路二：

def is_subtree(s: 'TreeNode', t: 'TreeNode') -> 'bool':

    def is_same(s, t):
        if s and t:
            equal = s.val==t.val
            if not t.left and not t.right:
                return equal
            else:
                return (equal and is_same(s.left, t.left) and
                        is_same(s.right, t.right))
        else:
            return s is t
    stack = s and [s]
    while stack:
        node = stack.pop()
        if node:
            res = is_same(node, t)
            if res:
                return True
            stack.append(node.right)
            stack.append(node.left)
    return False