102. Binary Tree Level Order Traversal

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

Example 1:

Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]] Example 2:

Input: root = [1] Output: [[1]] Example 3:

Input: root = [] Output: []

Constraints:

The number of nodes in the tree is in the range [0, 2000]. -1000 <= Node.val <= 1000

solving this problem requires a queue and we will store root and loop till q is not empty and each iteration we will add the left and right nodes of the current node and store the current nodes in a vector this way we will get the Level order traversal.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        if(!root) return {};
        vector<vector<int>> ans;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty())
        {
            int size = q.size();
            vector<int> temp;
            for(int i=0;i<size;i++)
            {
                TreeNode* t = q.front();
                q.pop();
                temp.push_back(t->val);
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
            }
            ans.push_back(temp);
        }
        return ans;        
    }
};