leetcode-107. Binary Tree Level Order Traversal II
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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
思路：
首先定义一个vector<vector<TreeNode*>> temp来保存二叉树的层次遍历结果，然后将temp中的层次中节点的值倒序赋值给vector<vector<int>> result。
accepted code:
1 /** 2 * Definition for a binary tree node. 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 vector<vector<int>> result; 14 vector<vector<TreeNode*>> temp; 15 if(root==nullptr) 16 return result; 17 vector<TreeNode*> tk; 18 tk.push_back(root); 19 temp.push_back(tk); 20 int i=0; 21 while(temp[i].size()>0) 22 { 23 tk.clear(); 24 for(int j=0;j<temp[i].size();j++) 25 { 26 if(temp[i][j]->left!=nullptr) 27 tk.push_back(temp[i][j]->left); 28 if(temp[i][j]->right!=nullptr) 29 tk.push_back(temp[i][j]->right); 30 } 31 temp.push_back(tk); 32 i++; 33 } 34 vector<int> num; 35 for(int i=temp.size()-2;i>=0;i--) 36 { 37 for(int j=0;j<temp[i].size();j++) 38 { 39 num.push_back(temp[i][j]->val); 40 } 41 result.push_back(num); 42 num.clear(); 43 } 44 return result; 45 } 46 };
