[Leetcode] 993. Cousins in Binary Tree

方法1

KEY POINT: if the index of root is n, then index of root.left is 2 * n, index of root.right is 2 * n + 1

BFS
use dict to store the node level and node index, check if they are in the same level and not siblings
Time complexity: O(n)
Space complexity: O(n)

class Solution:
    def isCousins(self, root: TreeNode, x: int, y: int) -> bool:
        if not root:
            return False
        
        index = {
     root.val: 1}
        levels = {
     root.val: 1}
        queue = deque([root])
        lx, ly = -1, -1
        while queue:
            for i in range(len(queue)):
                node = queue.popleft()
                if node.left:
                    queue.append(node.left)
                    index[node.left.val] = 2 * index[node.val]
                    levels[node.left.val] = levels[node.val] + 1
                if node.right:
                    queue.append(node.right)
                    index[node.right.val] = 2 * index[node.val] + 1
                    levels[node.right.val] = levels[node.val] + 1
                    
        if levels[x] == levels[y] and (index[x] // 2 != index[y] // 2):
            return True
        return False