**The 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 = 1Output:3Explanation: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 = 4Output:5Explanation: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 = 2Output:1

**Constraints:**

- The number of nodes in the tree is in the range
`[2, 10`

.^{5}] `-10`

^{9}<= Node.val <= 10^{9}- All
`Node.val`

are**unique**. `p != q`

`p`

and`q`

will exist in the tree.

**The Solution:**

var lowestCommonAncestor = function(root, p, q) { if(!root || root== p || root == q) return root let left = lowestCommonAncestor(root.left,p,q) let right = lowestCommonAncestor(root.right,p,q) return (left && right) ? root : (left || right) };