题解 | #在二叉树中找到两个节点的最近公共祖先#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param root TreeNode类 
 * @param o1 int整型 
 * @param o2 int整型 
 * @return int整型
 */
int lowestCommonAncestor(struct TreeNode* root, int o1, int o2 ) {
    // write code here
    if(root == NULL) return -1;
    if(root->val == o1 || root->val == o2) return root->val;
    int left = lowestCommonAncestor(root->left, o1, o2);
    int right = lowestCommonAncestor(root->right, o1, o2);
    if(left == -1) return right;
    if(right == -1) return left;
    return root->val;
}