题目：原题链接（困难）
标签：树、二叉树、动态规划
| 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) | 896ms (94.74%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class Solution:
# 对于每个三叉节点，至少有两边需要放导航
def __init__(self):
self.ans = 0
def navigation(self, root: TreeNode) -> int:
left = self.dfs(root.left)
right = self.dfs(root.right)
if left == 0 and right == 0:
return 1
elif left == 2 or right == 2:
return self.ans
elif left == 1 and right == 1:
return self.ans
else:
return self.ans + 1
def dfs(self, node):
"""返回节点的上一个三叉路，有几个放置的导航"""
if not node:
return 0
left = self.dfs(node.left)
right = self.dfs(node.right)
if node.left and node.right:
if left == 0 and right == 0:
self.ans += 1
return 1
elif left == 0 or right == 0:
return 1
else:
return 2
elif node.left:
return left
elif node.right:
return right
else:
return 0
