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.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int> > result;
        if(!root) return result;
        traverse(root, 0, result);
        for(int i = 0, j = result.size() - 1; i < j; i++, j--){
            swap(result[i], result[j]);
        }
        return result;
    }
    void traverse(TreeNode* root, int level, vector<vector<int> >& result){
        if(!root) return;
        if(level == result.size()) //current level does not exsit, we need to create it
            result.push_back(vector<int> ());
        
        result[level].push_back(root->val); //push the node in the level
        traverse(root->left, level + 1, result);
        traverse(root->right, level + 1, result);
    }
};

2. Iteration.

 　Runtime: 8ms.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int> >result;
        if(!root) return result;
        queue<TreeNode* > qu;
        qu.push(root);
        
        while(!qu.empty()){
            int n = qu.size();
            vector<int> level; //store the nodes value in current level
            while(n--){
                TreeNode* temp = qu.front();
                level.push_back(temp->val);
                qu.pop();
                
                if(temp->left) qu.push(temp->left);
                if(temp->right) qu.push(temp->right);
            }
            result.push_back(level);
        }
        
        for(int i = 0, j = result.size() - 1; i < j; i++, j--) 
            swap(result[i], result[j]);
        
        return result;
    }
};