已经正式在实习了，好久都没有刷题了（应该有半年了吧），感觉还是不能把思维锻炼落下，所以决定每周末刷一次LeetCode。
这是第一周（菜的真实，只做了两题，还有半小时不想看了，冷~）。
第一题：
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:
- The number of nodes in the given tree will be in the range
[1, 100]. - Each node's value will be an integer in the range
[0, 99].
题目意思很简单，就是给你一棵树，让你判断这棵树所有节点的值是不是都是同一个数。
直接遍历节点，然后记录下来再判断就好。（其实可以边遍历边判断）
/**
* 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 {
private:
int a[];
public: void view(TreeNode* root) {
if( root != NULL ) a[root->val] ++;
if( root->right != NULL ) view(root->right);
if( root->left != NULL ) view(root->left);
} bool isUnivalTree(TreeNode* root) {
memset(a, , sizeof(a));
view(root);
int cnt = ;
for(int i=; i<; i++) {
if( a[i] != ) cnt ++;
}
return cnt == ;
}
};
第二题：
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]
Note:
1 <= N <= 90 <= K <= 9
题目意思很简单，看样例基本能明白，给你一个长度n，和一个限定差值k，让你找出所有长度为n并且相邻数位之间的差值等于k的这些数（任何顺序），除0之外不能有任何数是以0开头。
有两个坑点：
1、当N为1的时候，0是正确的数。
2、当K为0的时候，注意不要重复计算。
class Solution {
public:
vector<int> numsSameConsecDiff(int N, int K) {
vector<int> ans;
if( N == ) ans.push_back();
for(int i=; i<; i++) {
queue<int> q;
q.push(i);
int len = N-;
while( len!= ) {
int si = q.size();
while( si -- ) {
int st = q.front(); q.pop();
int last = st % ;
if( last + K < ) q.push(st*+last+K);
if( last - K >= && (last+K != last-K) ) q.push(st*+(last-K));
}
len --;
}
while( !q.empty() ) {
int top = q.front();
ans.push_back(top);
q.pop();
}
}
return ans;
}
};
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