Solution.py

"""

Given the root of a binary tree, return the lowest common ancestor of its deepest leaves.
Recall that:

The node of a binary tree is a leaf if and only if it has no children
The depth of the root of the tree is 0. if the depth of a node is d, the depth of each of its children is d + 1.
The lowest common ancestor of a set S of nodes, is the node A with the largest depth such that every node in S is in the subtree with root A.

Note: This question is the same as 865: https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/
 
Example 1:


Input: root = [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation: We return the node with value 2, colored in yellow in the diagram.
The nodes coloured in blue are the deepest leaf-nodes of the tree.
Note that nodes 6, 0, and 8 are also leaf nodes, but the depth of them is 2, but the depth of nodes 7 and 4 is 3.
Example 2:

Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree, and it's the lca of itself.

Example 3:

Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest leaf node in the tree is 2, the lca of one node is itself.

 
Constraints:

The number of nodes in the tree will be in the range [1, 1000].
0 <= Node.val <= 1000
The values of the nodes in the tree are unique.


"""


# 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 lcaDeepestLeaves(self, root: TreeNode) -> TreeNode:
        def subtree_with_deepest(node):
            if node is None:
                return node, 0
            lt, ll = subtree_with_deepest(node.left)
            rt, rl = subtree_with_deepest(node.right)
            if ll == rl:
                return node, ll + 1
            elif ll > rl:
                return lt, ll + 1
            else:
                return rt, rl + 1
        
        return subtree_with_deepest(root)[0]