[Leetcode] 543. Diameter of Binary Tree_Easy Tag: DFS

Given the root of a binary tree, return 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.

The length of a path between two nodes is represented by the number of edges between them.

 

Example 1:

Input: root = [1,2,3,4,5]
Output: 3
Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].

Example 2:

Input: root = [1,2]
Output: 1

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -100 <= Node.val <= 100

 

Ideas: 最长一条边的node的个数 - 1， 建立一个helper function， get the max number of nodes with the root, 再用self.an = max(self.ans, left + right + 1 - 1) 去得到边

Code

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    
    def diameterOfBinaryTree(self, root: TreeNode) -> int:
        self.ans = 0
        self.helperWithRoot(root)
        return self.ans
    
    # get the number of nodes 
    def helperWithRoot(self, root):
        if not root: return 0
        left = self.helperWithRoot(root.left)
        right = self.helperWithRoot(root.right)
        self.ans = max(self.ans, left + right )
        return 1 + max(left, right)

 

get all the nodes, and then self.ans records the most nodes of one path, finally we decrease one for the ans.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def __init__(self):
        self.ans = 0
    
    def diameterOfBinaryTree(self, root: TreeNode) -> int:
        self.depth(root)
        return self.ans - 1
    
    def depth(self, root):
        if not root: return 0
        left = self.depth(root.left)
        right = self.depth(root.right)
        local = 1 + max(left, right)
        self.ans = max(self.ans, 1 + left + right)
        return local