1218 Lowest Common Ancestor Of Deepest Leaves

Lowest Common Ancestor of Deepest Leaves

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.

 

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**.

 

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

# 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: Optional[TreeNode]) -> Optional[TreeNode]:
        # Step 1: Level-order traversal to get the deepest level nodes
        lvl = [root]

        def test(lvl):
            temp = []
            prev = []
            while lvl:
                r = lvl.pop()
                prev.append(r)
                if r.left is not None:
                    temp.append(r.left)
                if r.right is not None:
                    temp.append(r.right)
            if temp:
                return test(temp)
            else:
                return prev

        prev = test(lvl)  # deepest leaves
        prev_set = set(prev)  # for faster lookup
        total = len(prev_set)
        res = None

        # Step 2: DFS to find LCA of deepest leaves
        def dfs(node):
            nonlocal res
            if node is None:
                return 0
            left = dfs(node.left)
            right = dfs(node.right)
            self_count = 1 if node in prev_set else 0
            count = left + right + self_count
            if count == total and res is None:
                res = node
            return count

        dfs(root)
        return res