# Breadth-first search | Binary Tree Level Order Traversal

Gowthaman Ravi
·Jan 28, 2022·

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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

Java Solution

`````` public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}

public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {

if(root == null) return wrapList;

queue.offer(root);
while(!queue.isEmpty()){
int levelNum = queue.size();
for(int i=0; i<levelNum; i++) {
if(queue.peek().left != null) queue.offer(queue.peek().left);
if(queue.peek().right != null) queue.offer(queue.peek().right);
}
}
return wrapList;
}
}
``````

C++ Solution

``````/**
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) {
vector<vector<int>> res;
vector<int> res1;
queue<TreeNode*> q;
TreeNode* t=NULL;
if(root==NULL)
return res;
q.push(root);

while(!q.empty()){
int s = q.size();

for( int i=0;i<s;i++){
t=q.front();
q.pop();
if(t->left!=NULL)
q.push(t->left);
if(t->right!=NULL)
q.push(t->right);
res1.push_back(t->val);
}
res.push_back(res1);
res1.resize(0);
}
return res;
}
};
``````