题解 | #在二叉树中找到两个节点的最近公共祖先#后序遍历

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
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
        if(root == nullptr) 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 && right != -1){
            return root->val;
        }
        if(left == -1 && right != -1){
            return right;
        }
        else if (left != -1 && right == -1) {
            return left;
        }
        else {
            return -1;
        }
    }
};