LeetCode题解：0100（相同的树）
题目链接（简单）
| 解法 | 执行用时 |
|---|---|
| Ans 1 (Python) | 40ms (>68.56%) |
| Ans 2 (Python) | 40ms (>68.56%) |
解法一（暴力解树）：
def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
if p is None and q is None:
return True
if (p is not None and q is not None) != (p is not None or q is not None):
return False
node_list = [(p, q)]
while node_list is not None:
pp, qq = node_list[0]
if pp.val != qq.val:
return False
if (pp.left is not None and qq.left is not None) != (pp.left is not None or qq.left is not None):
return False
if (pp.right is not None and qq.right is not None) != (pp.right is not None or qq.right is not None):
return False
if pp.left:
node_list.append((pp.left, qq.left))
if pp.right:
node_list.append((pp.right, qq.right))
node_list.pop(0)
return True
解法二（递归解树）：
def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
if p is None and q is None:
return True
if p is None or q is None:
return False
if p.val != q.val:
return False
return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
