0236. Lowest Common Ancestor of a Binary Tree

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree

Description

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Example 1:

**Input:** root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
**Output:** 3
**Explanation:** The LCA of nodes 5 and 1 is 3.

Example 2:

**Input:** root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
**Output:** 5
**Explanation:** The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

**Input:** root = [1,2], p = 1, q = 2
**Output:** 1

Constraints:

• The number of nodes in the tree is in the range [2, 105].

• -109 <= Node.val <= 109

• All Node.val are unique.

• p != q

• p and q will exist in the tree.

ac

Like bubble, surface up the p or q node.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) return null;

        boolean currIs = root == p || root == q;

        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        if (left == null && right == null && !currIs) {
            return null;
        } else if (currIs || (left != null && right != null) ) {
            return root;
        } else {
            return left == null ? right : left;
        }
    }
}

This one is cleaner, but they are the same. previous one is more natural in an interview.

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || root == p || root == q) return root;

        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        if (left != null && right != null) {
            return root;
        } else if (left != null) {
            return left;
        } else {
            return right;
        }
    }
}