May Leetcode Challenge

Problem

In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

Note:

The number of nodes in the tree will be between 2 and 100.
Each node has a unique integer value from 1 to 100.

Solution

Use level traverse technique
Can solve by using and level array to track level

class Solution {
    func isCousins(_ root: TreeNode?, _ x: Int, _ y: Int) -> Bool {
        if root == nil { return false }
        var levels : [[TreeNode]] = [[root!]]
        
        var xParent: TreeNode?
        var yParent: TreeNode?
        
        var depth = 0
        
        while depth < levels.count {
            var new = [TreeNode]()
            
            for node in levels.last! {
                if let left = node.left {
                    if x == left.val { xParent = node }
                    if y == left.val { yParent = node }
                    new.append(left)
                }
                
                if let right = node.right {
                    if x == right.val { xParent = node }
                    if y == right.val { yParent = node }
                    new.append(right)
                }
                
                if let xParent = xParent, let yParent = yParent, xParent.val != yParent.val {
                    return true
                }
            }
            
            if !new.isEmpty {
                levels.append(new)
            }
            
            xParent = nil
            yParent = nil
            depth += 1
        }
        
        return false
    }
}