题解 | #在二叉树中找到两个节点的最近公共祖先#

struct TreeNode* low(struct TreeNode* root, int o1, int o2) {
    if (root == NULL || root->val == o1 || root->val == o2) {
        return root;
    }
    struct TreeNode* left = low(root->left, o1, o2);
    struct TreeNode* right = low(root->right, o1, o2);
    if (left != NULL && right != NULL) {
        return root;
    }
    return left != NULL ? left : right;
}
int lowestCommonAncestor(struct TreeNode* root, int o1, int o2 ) {
    return low(root,o1,o2)->val;
}