1123. Lowest Common Ancestor of Deepest Leaves

Problem Statement

1123. Lowest Common Ancestor of Deepest Leaves

Medium

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/

Java

Source file

class Solution {

    public TreeNode lcaDeepestLeaves(TreeNode root) {
        return dfs(root).getKey();
    }

    private Pair<TreeNode, Integer> dfs(TreeNode root) {
        if (root == null) {
            return new Pair<>(null, 0);
        }

        Pair<TreeNode, Integer> left = dfs(root.left);
        Pair<TreeNode, Integer> right = dfs(root.right);

        if (left.getValue() > right.getValue()) {
            return new Pair<>(left.getKey(), left.getValue() + 1);
        }
        if (left.getValue() < right.getValue()) {
            return new Pair<>(right.getKey(), right.getValue() + 1);
        }
        return new Pair<>(root, left.getValue() + 1);
    }
}