难度:简单
给定一个二叉树,返回其节点值自底向上的层次遍历。 (即按从叶子节点所在层到根节点所在的层,逐层从左向右遍历)
例如:
给定二叉树 [3,9,20,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
返回其自底向上的层次遍历为:
[ [15,7], [9,20], [3] ]
题目分析:
使用双队列进行奇偶层次遍历,进行从上到下存储,存完之后后面再进行前后交换即可。
参考代码:
/**
* 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) {
vector<vector<int> > res;
if(root == NULL)
return res;
queue<TreeNode*> odd_queue;
queue<TreeNode*> even_queue;
vector<int> temp;
TreeNode* pMove = root;
odd_queue.push(root);
while(!odd_queue.empty() || !even_queue.empty())
{
while(!odd_queue.empty())
{
pMove = odd_queue.front();
odd_queue.pop();
temp.push_back(pMove->val);
if(pMove->left)
even_queue.push(pMove->left);
if(pMove->right)
even_queue.push(pMove->right);
}
if(!temp.empty())
{
res.push_back(temp);
temp.clear();
}
while(!even_queue.empty())
{
pMove = even_queue.front();
even_queue.pop();
temp.push_back(pMove->val);
if(pMove->left)
odd_queue.push(pMove->left);
if(pMove->right)
odd_queue.push(pMove->right);
}
if(!temp.empty())
{
res.push_back(temp);
temp.clear();
}
}
for(int i = 0; i < (res.size()>>1); i++)
{
swap(res[i],res[res.size()-1-i]);
}
return res;
}
};