43.二叉树的最近公共祖先

题目描述：

代码实现：

• 递归法，对左右子树分别递归，找到p，q后返回，若left，right同时找到pq，就返回根节点，如果只有left和right其中一个找到，就返回他们本身。

/**
 * 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 === null || root === p || root === q) {
        return root
    }
    var left = lowestCommonAncestor(root.left, p, q) 
    var right = lowestCommonAncestor(root.right, p, q)
    if (left && right) {
        return root
    } else if (left) {
        return left
    } else if (right) {
        return right
    }
};