这个算法的运行时间复杂度是多少?你是如何分析的

-下面的lowestCommonAncestor函数在二进制树中查找两个节点p和q的最低公共祖先(假设两个节点都存在并且所有节点值都是唯一的(。

class Solution {
public:
bool inBranch(TreeNode* root, TreeNode* node)
{
if (root == NULL)
return false;
if (root->val == node->val)
return true;
return (inBranch(root->left, node) || inBranch (root->right, node));
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (root == NULL)
return NULL;
if (root->val == p->val || root->val == q->val)
return root;
bool pInLeftBranch = false;
bool qInLeftBranch = false;
pInLeftBranch = inBranch (root->left, p);
qInLeftBranch = inBranch (root->left, q);
//Both nodes are on the left
if (pInLeftBranch && qInLeftBranch)
return lowestCommonAncestor(root->left, p, q);
//Both nodes are on the right
else if (!pInLeftBranch && !qInLeftBranch)
return lowestCommonAncestor(root->right, p, q);
else 
return root;
}
};