LeetCode-in-All
543. Diameter of Binary Tree
Easy
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
Solution
/**
* Definition for a binary tree node.
* public class TreeNode {
* public var val: Int
* public var left: TreeNode?
* public var right: TreeNode?
* public init() { self.val = 0; self.left = nil; self.right = nil; }
* public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
* public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
* self.val = val
* self.left = left
* self.right = right
* }
* }
*/
class Solution {
private var diameter: Int = 0
func diameterOfBinaryTree(_ root: TreeNode?) -> Int {
diameter = 0
_ = diameterOfBinaryTreeUtil(root)
return diameter
}
private func diameterOfBinaryTreeUtil(_ root: TreeNode?) -> Int {
guard let root = root else {
return 0
}
let leftLength = root.left != nil ? 1 + diameterOfBinaryTreeUtil(root.left) : 0
let rightLength = root.right != nil ? 1 + diameterOfBinaryTreeUtil(root.right) : 0
diameter = max(diameter, leftLength + rightLength)
return max(leftLength, rightLength)
}
}