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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,#,
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int> > levelOrder(TreeNode *root) { 13 vector<vector<int> > result; 14 if (root == nullptr) return result; 15 16 queue<TreeNode *> current, next; 17 vector<int> level; 18 19 current.push(root); 20 while (!current.empty()) { 21 while (!current.empty()) { 22 TreeNode *p = current.front(); 23 current.pop(); 24 level.push_back(p->val); 25 if (p->left != nullptr) next.push(p->left); 26 if (p->right != nullptr) next.push(p->right); 27 } 28 result.push_back(level); 29 level.clear(); 30 swap(current, next); 31 } 32 33 return result; 34 } 35 };
#,15,7},
3
/ 9 20
/ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
思路：用两个vector分层次记录值，时间复杂度O(n)，空间复杂度O(n)
[LeetCode] Binary Tree Level Order Traversal
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原文地址：http://www.cnblogs.com/vincently/p/4229728.html
