Lowest Common Ancestor of a Binary Tree

Problem

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.

Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function(root, p, q) {
    if (!root || root === p || root === q) {
        return root;
    } else {
        let l = lowestCommonAncestor(root.left, p, q);
        let r = lowestCommonAncestor(root.right, p, q);
        if (l && r) {
            return root;
        } else if (l) {
            return l;
        } else {
            return r;
        }
    }
};

We will implement a DFS solution. We traverse the entire tree and either return null if p and q are not found, or return the node of p and q if they are found. The LCA of p and q will be the node where both the left and right recursive calls return a non null value. Thus, return the current node if p and q are non-null. Otherwise, return the single non-null value from the recursive call (or null if both calls returned null).

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Problem

Solution