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/********************************************************************************** * * 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.* * OJ's Binary Tree Serialization:* * The serialization of a binary tree follows a level order traversal, where '#' signifies * a path terminator where no node exists below.* * Here's an example:* * 1* / \* 2 3* /* 4* \* 5* * The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}". * * **********************************************************************************/ /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ //对于这个问题需要注意的是如何给分层 //下面一种方法非常的巧妙，它用了last指针指向最后一个位置。class Solution {public: vector > levelOrder(TreeNode *root) { vector< vector > vv; if(root == NULL) return vv; int level = 0; // current level. TreeNode *last = root; // last node of currrent level. queue q; q.push(root); vv.push_back(vector ()); while(!q.empty()) { TreeNode *p = q.front(); q.pop(); vv[level].push_back(p->val); if(p->left ) q.push(p->left); if(p->right) q.push(p->right); if(p == last) { level++; last = q.back(); vv.push_back(vector ()); // new buffer for next row. } } vv.pop_back(); return vv;}}; //关于第二个不做讨论
//只需要用reserves(vv.begin(),vv,end())即可。转载地址：http://pvdoi.baihongyu.com/