题目:原题链接(简单)
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
---|---|---|---|
Ans 1 (Python) | O ( N ) O(N) O(N) | O ( N ) O(N) O(N) | 36ms (90.72%) |
Ans 2 (Python) | O ( n ) O(n) O(n) | O ( H ) O(H) O(H) | 40ms (77.25%) |
Ans 3 (Python) |
LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。
解法一(迭代):
def isUnivalTree(self, root: TreeNode) -> bool:
val = root.val
now_node = [root]
while now_node:
next_node = []
for node in now_node:
if node.val != val:
return False
if node.left:
next_node.append(node.left)
if node.right:
next_node.append(node.right)
now_node = next_node
return True
解法二(递归):
def __init__(self):
self.val = None
def isUnivalTree(self, root: TreeNode) -> bool:
self.val = root.val
def helper(node):
if not node:
return True
elif node.val != self.val:
return False
else:
return helper(node.left) and helper(node.right)
return helper(root)