本题经过一下午的思考,终于解出来了。使用的是层次遍历的思想。

class Solution {
public:
    bool isToeplitzMatrix(vector<vector<int>>& matrix) {
        int RowLen = matrix.size() - 1;
        int ColLen = matrix[0].size() - 1;
        int N = RowLen + ColLen;
        int i = RowLen;
        int j = 0;
        queue<pair<int, int>> Q;

        Q.push(make_pair(i, j));
        while (!Q.empty())
        {
            //全部出队,加入vector
            vector<pair<int, int>> V;
            while (!Q.empty())
            {
                pair<int, int> p = Q.front();
                Q.pop();
                V.push_back(p);
                cout << p.first << " " << p.second << endl;
            }
            cout << "------------------------------------------" << endl;

            //遍历V
            int base = matrix[V[0].first][V[0].second];
            set<pair<int, int>> S;
            for (auto v : V)
            {
                //判断是否都是相同的数值
                if (base != matrix[v.first][v.second])
                {
                    return false;
                }

                //判断“上”和“右”的元素是否合法,
                int Up_x = v.first - 1;
                int Up_y = v.second;
                //“上元素”合法则加入S(去重)
                if (Up_x >= 0 && S.find(make_pair(Up_x, Up_y)) == S.end())
                {
                    S.insert(make_pair(Up_x, Up_y));
                }

                int Right_x = v.first;
                int Right_y = v.second + 1;
                //“右元素”合法则加入S(去重)
                if (Right_y <= ColLen && S.find(make_pair(Right_x, Right_y)) == S.end())
                {
                    S.insert(make_pair(Right_x, Right_y));
                }
            }

            //将S中的元素,添加到Q中
            for (auto s : S)
            {
                Q.push(s);
            }
        }
        return true;
    }
};