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) {}
 * };
 */
#include <vector>
class Solution {
  public:
    void getPath(std::vector<TreeNode*>& path, TreeNode* root, int target) {
        // 深度优先搜索
        if (root == nullptr) {
            return;
        }
        path.push_back(root);
        if (path.back()->val != target) {
            getPath(path, root->left, target);
            getPath(path, root->right, target);
        }
        if (path.back()->val != target){
            path.pop_back();
        }
    }
    /**
     * 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
        std::vector<TreeNode*> path1, path2;
        getPath(path1, root, o1);
        getPath(path2, root, o2);
        int ret;
        for (int i = 0; i < path1.size() && i< path2.size(); i++) {
            if (path1[i]->val == path2[i]->val) {
                ret = path1[i]->val;
            } else {
                break;
            }
        }
        return ret;
    }
};