Given 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.
許久未練, 生疏不少. 歎~
/** * 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> levelOrder(TreeNode *root) { vector vee; vector >ve; queue qu; TreeNode *no; if(NULL==root) return ve; qu.push(root); int len = qu.size(); while(len!=0) { while(len--) { no = qu.front(); qu.pop(); vee.push_back(no->val); if(NULL!=no->left)qu.push(no->left); if(NULL!=no->right)qu.push(no->right); } ve.push_back(vee); vee.clear(); len = qu.size(); } return ve; } };