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 {
public:
    int flag = 0;

    int lowestCommonAncestor(TreeNode* root, int o1, int o2) {
        vector<int> path1, path2;

        dfs(root, path1, o1);
        flag = 0;
        dfs(root, path2, o2);

        int res = 0;
        for(int i=0; i<path1.size() && i<path2.size(); i++){
            if(path1[i] == path2[i]){  //共同祖先的之前的路径是相同的
                res = path1[i];
            }else{
                break;
            }
        }

        return res;
    }

    void dfs(TreeNode* root,vector<int>& path, int target){
        if(root==nullptr){
            return;
        }

        if(root->val == target){
            path.push_back(root->val);  //自己也可能是根节点，添加进去检测
            flag = 1;
            return;
        }

        path.push_back(root->val);

        dfs(root->left, path, target);
        if(flag == 1){
            return;
        }

        dfs(root->right, path, target);
        if( flag == 1 ){
            return;
        }

        path.pop_back();
        return;
    }
};