problem:
Given a binary tree, return thelevel ordertraversal 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.
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TreeBreadth-first Search
题意:层序遍历二叉树,每层要分开
thinking:
(1)普通的二叉树层序遍历算法每层是没有分开的,这道题要求分开。
(2)采用广度优先搜索(BFS)的思想
(3)采用两个queue,递归实现每层分开的层序遍历,很经典
code:
class Solution { private:vector<vector<int> > ret; public:vector<vector<int> > levelOrder(TreeNode *root) {ret.clear();if(root==NULL)return ret;queue<TreeNode *> tmp_queue;tmp_queue.push(root);level_order(tmp_queue);return ret;}void level_order(queue<TreeNode *> queue1){if(queue1.empty())return;vector<int> array;queue<TreeNode *> queue2;while(!queue1.empty()){TreeNode *tmp=queue1.front();array.push_back(tmp->val);queue1.pop();if(tmp->left!=NULL)queue2.push(tmp->left);if(tmp->right!=NULL)queue2.push(tmp->right);}ret.push_back(array);level_order(queue2);} };
,有的旅行时为了寻找逝去的年华,重温青春的惆怅。