Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / 
  9  20
    /  
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

Analyse: 

1. Recursion: If the level does not exist, create it and then push corresponding value into it. 

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>> levelOrder(TreeNode* root) {
        vector<vector<int> >result;
        traverse(root, 0, result);
        return result;
    }
    void traverse(TreeNode* root, int level, vector<vector<int> >& result){
        if(!root) return;
        if(level == result.size()) //the level does not exist and need to create it
            result.push_back(vector<int> ());
        
        result[level].push_back(root->val);
        traverse(root->left, level + 1, result);
        traverse(root->right, level + 1, result);
    }
};

2. Iteration: Using queue to store nodes. When poping the first node, we need to add its children(child) to the queue. Then keep poping until the queue is empty.

    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> > levelOrder(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 visited nodes in current level
            while(n--){
                TreeNode* temp = qu.front(); //pop the first node in the queue
                level.push_back(temp->val);
                qu.pop();
                if(temp->left) qu.push(temp->left); //add its children(child)
                if(temp->right) qu.push(temp->right);
            }
            result.push_back(level);
        }
        return result;
    }
};