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;
 *  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
  private:
    TreeNode* lowestCommonAncestorNode(TreeNode* root, int o1, int o2) {
        if (!root || root->val == o1 || root->val == o2) {
            return root;
        }
        TreeNode* left = lowestCommonAncestorNode(root->left, o1, o2);
        TreeNode* right = lowestCommonAncestorNode(root->right, o1, o2);
        if (left && right) {
            return root;
        }
        return left ? left : right;
    }

  public:
    /**
     * 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(TreeNode* root, int o1, int o2) {
        // write code here
        TreeNode* LCANode = lowestCommonAncestorNode(root, o1, o2);
        if (!LCANode) return -1;
        return LCANode->val;
    }
};