leetcode简单题6

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

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* 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.

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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; } };