代码实现：

思路：

递归判断左子树和右子树，查找p或者q是否在当前节点的子树上
1，在同一子树上，同一左子树，返回第一个找到的相同值，同一右子树上，返回第一个找到的相同值
2，不在同一子树是，说明当前节点就是他们的公共祖先节点

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     struct TreeNode *left;*     struct TreeNode *right;* };*/
struct TreeNode* lowestCommonAncestor(struct TreeNode *root, struct TreeNode *p, struct TreeNode *q) {if (p == root || q == root || root == NULL) {return root;}struct TreeNode *l = lowestCommonAncestor(root->left, p, q);struct TreeNode *r = lowestCommonAncestor(root->right, p, q);if (l != NULL && r != NULL) {return root;}if (l == NULL && r != NULL) {return r;} else if (l != NULL && r == NULL) {return l;} else {return NULL;}
}