Find Nearest Right Node in Binary Tree - LeetCode
Given the root of a binary tree and a node u in the tree, return the nearest node on the same level that is to the right of u__, or return__ null __if __u __is the rightmost node in its level__.
Example 1:
Input: root = [1,2,3,null,4,5,6], u = 4 Output: 5 Explanation: The nearest node on the same level to the right of node 4 is node 5.
Example 2:
Input: root = [3,null,4,2], u = 2 Output: null Explanation: There are no nodes to the right of 2.
Constraints:
The number of nodes in the tree is in the range [1, 105].
1 <= Node.val <= 105
All values in the tree are distinct.
u is a node in the binary tree rooted at root.
- code
class Solution:
def findNearestRightNode(self, root: TreeNode, u: TreeNode) -> Optional[TreeNode]:
level = [root]
while level:
for i, cur in enumerate(level):
if cur == u:
if i == len(level) - 1: return None
return level[i + 1]
level = [i for node in level for i in (node.left, node.right) if i]