113 lines
2.9 KiB
Markdown
113 lines
2.9 KiB
Markdown
---
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title: leetcode-543
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date: 2020-03-25 19:13:52
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tags:
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categories: leetcode
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---
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### 543. 二叉树的直径
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[题目](https://leetcode-cn.com/problems/diameter-of-binary-tree/)
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<!--more-->
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这题做出来了但是没有通过运行时间的测试，主要还是没想明白二叉树的直径到底是什么东西，用了个蠢办法。
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```python
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# Definition for a binary tree node.
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class TreeNode:
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def __init__(self, x):
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self.val = x
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self.left = None
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self.right = None
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class Solution:
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def diameterOfBinaryTree(self, root: TreeNode) -> int:
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if not root:
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return 0
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last_node = -1
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routes = []
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start = root
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node_stack = [root]
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while (start.left or start.right or node_stack):
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if start != node_stack[-1]:
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node_stack.append(start)
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if last_node == start.right:
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node_stack = node_stack[:-1]
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if not node_stack:
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break
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last_node = start
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start = node_stack[-1]
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continue
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if start.left and last_node != start.left:
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start = start.left
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last_node = start
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continue
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if start.right:
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start = start.right
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last_node = start
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continue
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routes.append(node_stack)
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node_stack = node_stack[:-1]
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if not node_stack:
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break
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last_node = start
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start = node_stack[-1]
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max_l = 0
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for route in routes:
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for route_ in routes:
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intersection = 0
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if route != route_:
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intersection = len(set(route).intersection(set(route_)))
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if intersection:
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intersection -= 1
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max_l = max(max_l, len(set(route)| set(route_)) - intersection)
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return max_l - 1
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```
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L43 之前做的是以深度优先的方式遍历一遍树，得出每个点的路径。后面的是将所有路径组合在一起得出任意两个点间的路径，算出最大长度。
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其实以某个点为根节点的树的直径，就是某个节点的**左子树的深度和右子树的深度的和**，用递归来处理这个会比较容易理解
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```python
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class Solution(object):
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def diameterOfBinaryTree(self, root):
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self.ans = 1
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def depth(node):
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# 访问到空节点了，返回0
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if not node: return 0
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# 左儿子为根的子树的深度
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L = depth(node.left)
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# 右儿子为根的子树的深度
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R = depth(node.right)
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# 计算d_node即L+R+1 并更新ans
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self.ans = max(self.ans, L+R+1)
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# 返回该节点为根的子树的深度
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return max(L, R) + 1
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depth(root)
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return self.ans - 1
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作者：LeetCode-Solution
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链接：https://leetcode-cn.com/problems/diameter-of-binary-tree/solution/er-cha-shu-de-zhi-jing-by-leetcode-solution/
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来源：力扣（LeetCode）
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著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
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```
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