Problem Statement

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

**Example :**Given binary tree

`3

/ \ 9 20 / \ 15 7`

return its level order traversal as:

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

Also think about a version of the question where you are asked to do a level order traversal of the tree when depth of the tree is much greater than number of nodes on a level.

Problem Link

Level Order - InterviewBit

Code (Using BFS)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
vector<vector<int> > Solution::levelOrder(TreeNode* root) {
    
    vector<vector<int>> res;
    
    if(!root)
    return res;
    
    queue<TreeNode*> q;
    q.push(root);
    
    while(!q.empty())
    {
        int lsize = q.size();
        vector<int> temp;
        while(lsize--)
        {
            TreeNode* curr = q.front();
            temp.push_back(curr->val);
            q.pop();
            
            if(curr->left)
            q.push(curr->left);
            
            if(curr->right)
            q.push(curr->right);
        }
        res.push_back(temp);
    }
    return res;
    
}