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Description of the problem
Given a binary tree, you need to calculate the length of its diameter. A binary tree is the maximum length of the diameter of any two nodes in the path lengths. This path may pass through the root.
Example:
given binary tree
where the code is inserted sheet
1
/ \
2 3
/ \
4 5
Return 3, which is the path length [4,2,1,3] or [5,2,1,3].
Note: the path length between two nodes is represented by the number of edges therebetween.
2 solution to a problem - recursion
Reference answers chiefs also encountered the same problem, start a direct count depth on both sides, then both sides of the root sum
class Solution:
def diameterOfBinaryTree(self, root: TreeNode) -> int:
if not root:
return 0
return self.get_height(root.left) + self.get_height(root.right)
def get_height(self, root):
if not root:
return 0
return max(self.get_height(root.left), self.get_height(root.right)) + 1
A wrong use cases:
[4,-7,-3,null,null,-9,-3,9,-7,-4,null,6,null,-6,-6,null,null,0,6,5,null,9,null,null,-1,-4,null,null,null,-2]
Visualization look like
The truth
I have a need to preserve the value of this every time I compare the updated value of the maximum diameter, with self.max 0 = to initialize the value
when the value obtained each time a node of the left subtree and right subtree, have need to compare the size of the height of the right subtree, save more self.max and left sub-tree height + down
and then ask how the left subtree and right subtree height of it, that is the height of the classic recursion problem: max ( depth (root.left), depth (root.right )) + 1
class Solution:
def __init__(self):
self.max_diameter=0
def diameterOfBinaryTree(self, root: TreeNode) -> int:
self.depth(root)
return self.max_diameter
def depth(self,root:TreeNode):
if not root:
return 0
l=self.depth(root.left)
r=self.depth(root.right)
self.max_diameter=max(self.max_diameter,l+r)
return max(l,r)+1
