今天的华师
Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
思路：
树的层次遍历，技巧是用NULL作为每一行的结尾
问题是什么时候才能是结尾呢？
当遇到队列中front是NULL时，表明上一行已经处理完了，则上一行的子节点也处理完了（处理每个节点的时候，让他的孩子节点入队列），此时把NULL加入队列
当队列中只有一个NULL的时候，则遍历完毕
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
|
* 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:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> result;
if (root == NULL)
return result;
vector<int> line;
queue<TreeNode*> Q;
Q.push(root);
Q.push(NULL);
while (Q.size()>1)
{
TreeNode *p = Q.front();
Q.pop();
if (p == NULL)
{
大专栏 leetcode简单题6> result.push_back(line);
line.clear();
Q.push(NULL);
}
else
{
line.push_back(p->val);
if (p->left != NULL)
{
Q.push(p->left);
}
if (p->right != NULL)
{
Q.push(p->right);
}
}
}
result.push_back(line);
reverse(result.begin(), result.end());
return result;
}
};
|
Plus One
Given a non-negative number represented as an array of digits, plus one to the number.
The digits are stored such that the most significant digit is at the head of the list.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
| class Solution {
public:
vector<int> plusOne(vector<int>& digits) {
reverse(digits.begin(),digits.end());
int len=digits.size();
int carry=1;
for(int i=0;i<len;i++)
{
int temp=digits[i]+carry;
if(temp==10)
{
digits[i]=0;
carry=1;
}
else
{
digits[i]=temp;
carry=0;
}
}
if(carry==1)
digits.push_back(1);
reverse(digits.begin(),digits.end());
return digits;
}
};
|