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] ]/** * 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> > result; if (root == NULL) { return result; } queue<TreeNode*> buf; buf.push(root); buf.push(NULL); vector<int> cur; while (buf.size()) { TreeNode *temp = buf.front(); if (temp != NULL) { cur.push_back(temp->val); if (temp->left) { buf.push(temp->left); } if (temp->right) { buf.push(temp->right); } } else { result.push_back(cur); cur.clear(); if (buf.size() == 1) { break; } else { buf.push(NULL); } } buf.pop(); } return result; } };