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 };