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 ans;
    bool helper(TreeNode* root, int o1, int o2){
        if(root == NULL) return false;
        bool isLeft = helper(root -> left, o1, o2);
        bool isRight = helper(root -> right, o1, o2);
        if((isLeft && isRight) 
           || (isLeft && (root -> val == o1 || root -> val == o2)) 
           || (isRight && (root -> val == o1 || root -> val == o2))) ans = root -> val;
        return isLeft || isRight || root -> val == o1 || root -> val == o2;
    }
    int lowestCommonAncestor(TreeNode* root, int o1, int o2) {
        // write code here
        helper(root, o1, o2);
        return ans;
    }
};