Binary Tree Level Order Traversal

Practice here

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).

Implementation: BFS (Breadth-First Search)#

• Intuition: A queue naturally enforces FIFO order, which guarantees that all nodes at depth d are processed before any node at depth d+1. Snapshotting the queue size at the start of each outer loop iteration tells you exactly how many nodes belong to the current level.
• Mechanics: The root is enqueued first. On each outer iteration, the current queue size is saved; that many nodes are dequeued in the inner loop, their values collected into a level vector, and their children enqueued. The level vector is then appended to the result.
• Trade-off: BFS is the canonical choice for level-order traversal because it visits nodes in the exact order required. The queue holds at most O(w) nodes at a time, where w is the maximum width of the tree — O(n) in the worst case for a complete tree's last level.

Visit all nodes at level 0 → then level 1 → then level 2 → …

Use a queue to process nodes level-by-level, pushing children as you go, so that all nodes at the same depth are visited together.

/**
 * 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>> traversalResult;
        queue<TreeNode*> q;
        q.push(root);

        while(!q.empty())
        {
            int size = q.size();
            vector<int> levelResult;
            for(int i=0;i<size;i++)
            {
                TreeNode* currNode = q.front();
                q.pop();

                levelResult.push_back(currNode->val);

                if(currNode->left)
                    q.push(currNode->left);

                if(currNode->right)
                    q.push(currNode->right);
            }
            traversalResult.push_back(levelResult);
        }
        return traversalResult;
    }
};

Time Complexity: O(n), Each node visited once

Space Complexity: O(n), the queue can store all the nodes of the last level.