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,#,#,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.
Analyse: same as the solutions in Binary Tree Level Order Traversal except we need to reverse the sequence in the vector.
1. Recursion.
Runtime: 4ms.
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 if(!root) return result; 15 traverse(root, 0, result); 16 for(int i = 0, j = result.size() - 1; i < j; i++, j--){ 17 swap(result[i], result[j]); 18 } 19 return result; 20 } 21 void traverse(TreeNode* root, int level, vector<vector<int> >& result){ 22 if(!root) return; 23 if(level == result.size()) //current level does not exsit, we need to create it 24 result.push_back(vector<int> ()); 25 26 result[level].push_back(root->val); //push the node in the level 27 traverse(root->left, level + 1, result); 28 traverse(root->right, level + 1, result); 29 } 30 };
2. Iteration.
Runtime: 8ms.
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 if(!root) return result; 15 queue<TreeNode* > qu; 16 qu.push(root); 17 18 while(!qu.empty()){ 19 int n = qu.size(); 20 vector<int> level; //store the nodes value in current level 21 while(n--){ 22 TreeNode* temp = qu.front(); 23 level.push_back(temp->val); 24 qu.pop(); 25 26 if(temp->left) qu.push(temp->left); 27 if(temp->right) qu.push(temp->right); 28 } 29 result.push_back(level); 30 } 31 32 for(int i = 0, j = result.size() - 1; i < j; i++, j--) 33 swap(result[i], result[j]); 34 35 return result; 36 } 37 };
