Lowest Common Ancestor of a Binary Tree

Descrition

doc

Solutions

First Idea

function oneOf(p, q, ref) {
  return ref.val === p.val || ref.val === q.val
}

function identical(p, q, left, right) {
  return (
    (left?.val === p.val && right?.val === q.val) ||
    (left?.val === q.val && right?.val === p.val)
  )
}

/**
 * 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) {
    return null
  }

  const left = lowestCommonAncestor(root.left, p, q)
  const right = lowestCommonAncestor(root.right, p, q)

  if (identical(p, q, left, right) || oneOf(p, q, root)) {
    return root
  }

  return left !== null ? left : right
}

• Time Complexity: O(n)
• Space Complexity: O(1)