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], ]
与[leetcode] Binary Tree Level Order Traversal的方法相同，利用一个队列来完成层次遍历，最后在把结果反转过来即可。
代码如下：
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> > levelOrderBottom(TreeNode *root) { 13 // Note: The Solution object is instantiated only once and is reused by each test case. 14 15 vector<vector<int>> result; 16 vector<vector<int>> result2; 17 if(root==NULL) return result; 18 vector<int> temp; 19 queue<TreeNode * > q; 20 q.push(NULL); 21 q.push(root); 22 while(q.size()>1) 23 { 24 if(q.front()==NULL) 25 { 26 q.pop(); 27 q.push(NULL); 28 if(temp.size()!=0) result.push_back(temp); 29 temp.clear(); 30 } 31 else 32 { 33 if(q.front()->left!=NULL)q.push(q.front()->left); 34 if(q.front()->right!=NULL)q.push(q.front()->right); 35 temp.push_back(q.front()->val); 36 q.pop(); 37 } 38 } 39 if(temp.size()!=0) result.push_back(temp); 40 for(int i = result.size() - 1; i >= 0; i--) result2.push_back(result.at(i)); 41 return result2; 42 43 } 44 };