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], ]
BFS solution:
1 vector<vector<int> > levelOrderBottom(TreeNode *root) { 2 vector<vector<int> > ret; 3 if(root == NULL) 4 return ret; 5 vector<TreeNode *> level; 6 level.push_back(root); 7 while(true) { 8 if(level.size() == 0) 9 break; 10 vector<int> nums; 11 vector<TreeNode *> tmp; 12 for(auto item : level) { 13 nums.push_back(item->val); 14 if(item->left != NULL) 15 tmp.push_back(item->left); 16 if(item->right != NULL) 17 tmp.push_back(item->right); 18 } 19 ret.insert(ret.begin(), nums); 20 level = tmp; 21 } 22 return ret; 23 }
DFS solution:
1 void getLevelNums(TreeNode *root, vector<vector<int> > &ret, int level) { 2 if(ret.size() < level + 1){ 3 vector<int> nums; 4 nums.push_back(root->val); 5 ret.insert(ret.begin(), nums); 6 }else if (ret.size() >= level + 1) { 7 ret[ret.size() - level - 1].push_back(root->val); 8 } 9 10 if(root->left != NULL) 11 getLevelNums(root->left, ret, level + 1); 12 13 if(root->right != NULL) 14 getLevelNums(root->right, ret, level + 1); 15 16 } 17 vector<vector<int> > levelOrderBottom(TreeNode *root) { 18 vector<vector<int> > ret; 19 if(root == NULL) 20 return ret; 21 getLevelNums(root, ret, 0); 22 return ret; 23 }