Leetcode-Challenges-Python/December_LeetCode_Challenge/Smallest_Subree_With_All_The_Deepest_Node.py at master · GSNCodes/Leetcode-Challenges-Python · GitHub

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Given the root of a binary tree, the depth of each node is the shortest distance to the root.
Return the smallest subtree such that it contains all the deepest nodes in the original tree.
A node is called the deepest if it has the largest depth possible among any node in the entire tree.
The subtree of a node is tree consisting of that node, plus the set of all descendants of that node.

Note: This question is the same as 1123:
https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/

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 nodes of the tree.
Notice that nodes 5, 3 and 2 contain the deepest nodes in the tree but node 2
is the smallest subtree among them, so we return it.

Example 2:
Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree.

Example 3:
Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest node in the tree is 2,
the valid subtrees are the subtrees of nodes 2, 1 and 0 but the subtree of node 2 is the smallest.


Constraints:
The number of nodes in the tree will be in the range [1, 500].
0 <= Node.val <= 500
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


# O(n) Time and O(n) Space
class Solution:
    def subtreeWithAllDeepest(self, root: TreeNode) -> TreeNode:

        def helper(node, depth=0):
            if node is None:
                return None, 0

            left_node, left_depth = helper(node.left, depth)
            right_node, right_depth = helper(node.right, depth)

            if left_depth == right_depth:
                return node, left_depth + 1

            elif left_depth < right_depth:
                return right_node, right_depth + 1

            else:
                return left_node, left_depth + 1

        return helper(root)[0]