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.
/** * 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<vector<int>> res; if(root == NULL) return res; int save; queue<TreeNode*> use; use.push(root); save = 1; while(save!=0) { int count = save; save = 0; vector<int> t; while(count > 0 && !use.empty()) { TreeNode* temp = use.front(); use.pop(); count--; t.push_back(temp->val); if(temp->left!=NULL) { use.push(temp->left); save++; } if(temp->right!=NULL) { use.push(temp->right); save++; } } res.push_back(t); } return res; } };
LeetCode--Binary Tree Level Order Traversal
原文地址:http://blog.csdn.net/shaya118/article/details/42706477