LeetCode OJ - Binary Tree Level Order Traversal

题目：

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]
]

解题思路：

广度优先遍历。

代码：

/**
* 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>> ans;
if (root == NULL) return ans;

queue<TreeNode *> one, another;
one.push(root);

while (!one.empty() || !another.empty()) {
if (!one.empty()) {
vector<int> cur_ans;
while (!one.empty()) {
TreeNode * node = one.front();
cur_ans.push_back(node->val);
if (node->left != NULL) another.push(node->left);
if (node->right != NULL) another.push(node->right);
one.pop();
}
ans.push_back(cur_ans);
}
if (!another.empty()) {
vector<int> cur_ans;
while (!another.empty()) {
TreeNode *node = another.front();
cur_ans.push_back(node->val);
if (node->left != NULL) one.push(node->left);
if (node->right != NULL) one.push(node->right);
another.pop();
}
ans.push_back(cur_ans);
}
}
return ans;
}
};