LeetCode [102] Binary Tree Level Order Traversal

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

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/** * Definition for a binary tree node. * 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>> ret; helper(ret, 0, root); return ret; } void helper(vector<vector<int>>& ret, int level, TreeNode* node){ if(!node) return; if(ret.size()<level+1){ ret.resize(level+1); } ret[level].push_back(node->val); helper(ret, level+1, node->left); helper(ret, level+1, node->right); } }; <script class="brush: js" type="syntaxhighlighter"><![CDATA[ /** * Definition for a binary tree node. * 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>> ret; if(!root) return ret; queue<TreeNode*> que, que_tmp; que.push(root); vector<int> layer; while(!que.empty()){ TreeNode *node = que.front(); layer.push_back(node->val); que.pop(); if(node->left) que_tmp.push(node->left); if(node->right) que_tmp.push(node->right); if(que.empty()){ ret.push_back(layer); layer.clear(); que = que_tmp; que_tmp = queue<TreeNode*>(); } } return ret; } };