LeetCode OJ - Binary Tree Level Order Traversal II

题目：

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree

{3,9,20,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

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

解题思路：

广度优先遍历，然后对结果进行翻转。

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> > levelOrderBottom(TreeNode *root) {
vector<vector<int> > ans;
if (root == NULL) return ans;
queue<TreeNode*> one;
queue<TreeNode*> another;
vector<int> cur_ans;
TreeNode* cur_node;

one.push(root);

while (!one.empty() || !another.empty()) {
if (!one.empty()) {
while (!one.empty()) {
cur_node = one.front();
one.pop();
cur_ans.push_back(cur_node->val);
if (cur_node->left) another.push(cur_node->left);
if (cur_node->right) another.push(cur_node->right);
}
ans.push_back(cur_ans);
cur_ans.clear();
}
if (!another.empty()) {
while (!another.empty()) {
cur_node = another.front();
another.pop();
cur_ans.push_back(cur_node->val);
if (cur_node->left) one.push(cur_node->left);
if (cur_node->right) one.push(cur_node->right);
}
ans.push_back(cur_ans);
cur_ans.clear();
}
}
reverse(ans.begin(), ans.end());
return ans;
}
};