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：
1 * 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int> > levelOrder(TreeNode *root) { 13 vector<vector<int>> result; 14 traverse(root,1,result); 15 std::reverse(result.begin(),result.end());//比层序遍历1多此一行 16 return result; 17 } 18 void traverse(TreeNode *root,size_t level,vector<vector<int>> &result) 19 { 20 if(root==nullptr) return; 21 if(level>result.size()) result.push_back(vector<int>()); 22 result[level-1].push_back(root->val); 23 traverse(root->left,level+1,result); 24 traverse(root->right,level+1,result); 25 } 26 };
二、迭代实现：
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 /*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*/ 11 class Solution { 12 public: 13 vector<vector<int> > levelOrder(TreeNode *root) { 14 vector<vector<int>> vec_vec_tree;//创建空vector,存放最后返回的遍历二叉树的值 15 vector<int> level;//创建空的vector，存放每一层的遍历二叉树的值 16 TreeNode *p=root; 17 if(p==nullptr) return vec_vec_tree;//如果二叉树空，返回空vector<vector<int>> 18 queue<TreeNode *> queue_tree;//创建一个空队列 19 queue_tree.push(p);//root节点入队列 20 queue_tree.push(nullptr);//空节点入队列 21 while(!queue_tree.empty())//直到队列为空 22 { 23 p=queue_tree.front(); //头结点取值，并出队列 24 queue_tree.pop(); 25 if(p==nullptr&&!level.empty())//节点为空并且队列不为空 26 { 27 queue_tree.push(nullptr);//入队空节点，与下一层隔开 28 vec_vec_tree.push_back(level);//已遍历的层入队 29 level.clear();//清空vecor level 30 } 31 else if(p!=nullptr)//如果节点不空 32 { 33 level.push_back(p->val);//遍历 34 if(p->left) queue_tree.push(p->left);//若有左右孩子，则入队列 35 if(p->right) queue_tree.push(p->right);//注意入队顺序：先左后右 36 } 37 38 } 39 std::reverse(vec_vec_tree.begin(),vec_vec_tree.end());//比层序遍历1多此一行 40 return vec_vec_tree; 41 } 42 };
b 迭代实现2
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 /*两个队列实现*/ 11 class Solution { 12 public: 13 vector<vector<int> > levelOrder(TreeNode *root) { 14 vector<vector<int> > result; 15 if(root == nullptr) return result; 16 queue<TreeNode*> current, next;//两个队列，保存当层（current）和下层（next）节点 17 vector<int> level; // 存放每层的节点值 18 current.push(root); 19 while (!current.empty())//直到二叉树遍历完成 20 { 21 while (!current.empty())//直到本层二叉树遍历完成 22 { 23 TreeNode* node = current.front();//取队首节点，出队列 24 current.pop(); 25 level.push_back(node->val);//访问节点值 26 if (node->left != nullptr) next.push(node->left);//若存在左右节点，则压入next队列中 27 if (node->right != nullptr) next.push(node->right);//注意入队顺序为先left后right 28 } 29 result.push_back(level);//压入总vector 30 level.clear();//清vector 31 swap(next, current);//next队列和current队列交换 32 } 33 std::reverse(result.begin(),result.end());//比层序遍历1多此一行 34 return result; 35 } 36 };