首页 > 代码库 > LeetCode--Binary Tree Level Order Traversal
LeetCode--Binary Tree Level Order Traversal
Binary Tree Level Order Traversal
Total Accepted: 12441 Total Submissions: 40879My SubmissionsGiven a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7}
,
3 / 9 20 / 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int> > levelOrder(TreeNode *root) { vector<int> vectorTemp; vector<vector<int> > vectorTotal; if(root==NULL)return vectorTotal; queue<TreeNode*> Q; TreeNode* p=root; Q.push(p); int counter=1; while(!Q.empty()) { p=Q.front(); Q.pop(); counter--; vectorTemp.push_back(p->val); if(p->left!=NULL)Q.push(p->left); if(p->right!=NULL)Q.push(p->right); if(counter==0) { //vector<int> vectorTemp; vectorTotal.push_back(vectorTemp); vectorTemp.clear(); counter=Q.size(); } } //vectorTotal.push_back(vectorTemp); return vectorTotal; } };
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。