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]
]
思路：使用两个队列完成层次搜索，结果存储在vector中。
若使用一个队列，也可完成层次遍历，只是不清楚遍历到哪个层次了。两个队列可以知道目前元素所在的层次。
/**
* 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>> levelOrderBottom(TreeNode* root) {
result.clear();
if(!root) return result; vector<int> level;
TreeNode * current;
queue<TreeNode * > queToPop;
queue<TreeNode * > queToPush;
queToPop.push(root);
while(!queToPop.empty())
{
while(!queToPop.empty())
{
current = queToPop.front();
queToPop.pop();
level.push_back(current->val);
if(current->left)
{
queToPush.push(current->left); //若使用一个队列，这里应该是queToPop.push(...)，只要queToPop不为空，就继续pop和push
}
if(current->right)
{
queToPush.push(current->right);
}
} result.push_back(level);
level.clear();
swap(queToPop, queToPush);
} reverse(result.begin(), result.end()); //逆序vector的方法
return result;
}
private:
vector<vector<int>> result;
};