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]
]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
说明：
1）与层序遍历1相似，只是本题遍历方向不同：从下往上，其他相同
2) 代码只要加上一行逆序输出的语句即可
实现：
一、 递归实现：
*
* 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;
traverse(root,,result);
std::reverse(result.begin(),result.end());//比层序遍历1多此一行
return result;
}
void traverse(TreeNode *root,size_t level,vector<vector<int>> &result)
{
if(root==nullptr) return;
if(level>result.size()) result.push_back(vector<int>());
result[level-].push_back(root->val);
traverse(root->left,level+,result);
traverse(root->right,level+,result);
}
};
二、迭代实现：
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/*常规层序遍历思想，每层入队结束压入一个空节点，作为标志*/
class Solution {
public:
vector<vector<int> > levelOrder(TreeNode *root) {
vector<vector<int>> vec_vec_tree;//创建空vector,存放最后返回的遍历二叉树的值
vector<int> level;//创建空的vector，存放每一层的遍历二叉树的值
TreeNode *p=root;
if(p==nullptr) return vec_vec_tree;//如果二叉树空，返回空vector<vector<int>>
queue<TreeNode *> queue_tree;//创建一个空队列
queue_tree.push(p);//root节点入队列
queue_tree.push(nullptr);//空节点入队列
while(!queue_tree.empty())//直到队列为空
{
p=queue_tree.front(); //头结点取值，并出队列
queue_tree.pop();
if(p==nullptr&&!level.empty())//节点为空并且队列不为空
{
queue_tree.push(nullptr);//入队空节点，与下一层隔开
vec_vec_tree.push_back(level);//已遍历的层入队
level.clear();//清空vecor level
}
else if(p!=nullptr)//如果节点不空
{
level.push_back(p->val);//遍历
if(p->left) queue_tree.push(p->left);//若有左右孩子，则入队列
if(p->right) queue_tree.push(p->right);//注意入队顺序：先左后右
} }
std::reverse(vec_vec_tree.begin(),vec_vec_tree.end());//比层序遍历1多此一行
return vec_vec_tree;
}
};
b 迭代实现2
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/*两个队列实现*/
class Solution {
public:
vector<vector<int> > levelOrder(TreeNode *root) {
vector<vector<int> > result;
if(root == nullptr) return result;
queue<TreeNode*> current, next;//两个队列，保存当层（current）和下层（next）节点
vector<int> level; // 存放每层的节点值
current.push(root);
while (!current.empty())//直到二叉树遍历完成
{
while (!current.empty())//直到本层二叉树遍历完成
{
TreeNode* node = current.front();//取队首节点，出队列
current.pop();
level.push_back(node->val);//访问节点值
if (node->left != nullptr) next.push(node->left);//若存在左右节点，则压入next队列中
if (node->right != nullptr) next.push(node->right);//注意入队顺序为先left后right
}
result.push_back(level);//压入总vector
level.clear();//清vector
swap(next, current);//next队列和current队列交换
}
std::reverse(result.begin(),result.end());//比层序遍历1多此一行
return result;
}
};