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;
 * };
 */

class Solution {
public:
    /**
     * 
     * @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* a = cal(root, o1, o2);
        return a->val;
    }
    
    TreeNode* cal(TreeNode* root, int o1, int o2){
        if(root == nullptr){
            return nullptr;
        }

        if(root->val == o1 || root->val == o2){
            return root;
        }
        
        TreeNode* left = cal(root->left, o1, o2);
        TreeNode* right = cal(root->right, o1, o2);
        
        if(left != nullptr && right != nullptr){
            return root;
        }
        
        if(left == nullptr && right == nullptr){
            return nullptr;
        }
        
        return left == nullptr ? right : left;
    }
    
};