题目描述:
LeetCode 543. Diameter of Binary Tree
Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.
Example:
Given a binary tree
1 / \ 2 3 / \ 4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
Note: The length of path between two nodes is represented by the number of edges between them.
题目大意:
给定一棵二叉树,计算任意两节点之间的边数的最大值。
解题思路:
解法I 计算子树深度
Python代码:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def traverse(self, root):
if not root: return 0
left = self.traverse(root.left)
right = self.traverse(root.right)
self.ans = max(self.ans, left + right)
return max(left, right) + 1
def diameterOfBinaryTree(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.ans = 0
self.traverse(root)
return self.ans
解法II 遍历二叉树 + 计算子树深度
Python代码:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def depth(self, root):
if not root: return 0
return 1 + max(self.depth(root.left), self.depth(root.right))
def traverse(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root: return 0
return max(self.depth(root.left) + 1 + self.depth(root.right), \
self.traverse(root.left), \
self.traverse(root.right))
def diameterOfBinaryTree(self, root):
"""
:type root: TreeNode
:rtype: int
"""
return max(self.traverse(root) - 1, 0)
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