Weekly Contest 117

965. Univalued Binary Tree

A binary tree is univalued if every node in the tree has the same value.

Return true if and only if the given tree is univalued.

Example 1:

Input: [1,1,1,1,1,null,1]
Output: true

Example 2:

Input: [2,2,2,5,2]
Output: false

Note:

1. The number of nodes in the given tree will be in the range [1, 100].
2. Each node‘s value will be an integer in the range [0, 99].

Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isUnivalTree(TreeNode* root) {
        solve(root, root->val);
        return ans;
    }

    void solve(TreeNode* root, int num) {
        if (root->val != num) ans = false;
        if (root->left != NULL) solve(root->left, num);
        if (root->right != NULL) solve(root->right, num);
    }

private:
    bool ans = true;
};

967. Numbers With Same Consecutive Differences

Return all non-negative integers of length N such that the absolute difference between every two consecutive digits is K.

Note that every number in the answer must not have leading zeros except for the number 0 itself. For example, 01 has one leading zero and is invalid, but 0 is valid.

You may return the answer in any order.

Example 1:

Input: N = 3, K = 7
Output: [181,292,707,818,929]
Explanation: Note that 070 is not a valid number, because it has leading zeroes.

Example 2:

Input: N = 2, K = 1
Output: [10,12,21,23,32,34,43,45,54,56,65,67,76,78,87,89,98]

Code:

class Solution {
public:
    vector<int> numsSameConsecDiff(int N, int K) {
        vector<int> ans;
        if (N == 1) {
            for (int i = 0; i < 10; ++i)
                ans.push_back(i);
            return ans;
        }
        for (int i = 1; i < 10; ++i) {
            string s = to_string(i);
            dfs(s, N, K, ans);
        }
        return ans;
    }

    void dfs(string s, const int N, const int K, vector<int>& ans) {
        if (s.length() == N) {
            ans.push_back(stoi(s));
            return ;
        }
        int lastNum = s[s.length()-1] - ‘0‘;
        int temp = lastNum + K;
        string dummy = s;
        if (temp >= 0 && temp < 10) {
            dummy += to_string(temp);
            dfs(dummy, N, K, ans);
        }
        if (K != 0) {
            int temp = lastNum - K;
            if (temp >= 0 && temp < 10) {
                s += to_string(temp);
                dfs(s, N, K, ans);
            }
        }
    }
};

原文地址：https://www.cnblogs.com/ruruozhenhao/p/10199215.html