[LeetCode] 102. 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,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]

Thought process:
Do a BFS starting from the root node. Because we need to offer nodes to the queue as we poll from the queue, we need to get the size of the queue at the beginning of the while loop, and only poll that many nodes from the queue.

Solution 1 (BFS):
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> levels = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        if (root != null) {
            queue.offer(root);
        }
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                level.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            levels.add(level);
        }
        
        return levels;
    }
}

Time complexity: O(n).

This problem can also be solved using DFS. We can do a DFS starting from the root node, and add nodes to the result list based on their heights.

Solution 2 (DFS):
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> levelOrder = new ArrayList<>();
        
        levelOrder(root, levelOrder, 0);
        
        return levelOrder;
    }
    
    private void levelOrder(TreeNode root, List<List<Integer>> levelOrder, int height) {
        if (root == null) {
            return;
        }
        
        if (height >= levelOrder.size()) {
            levelOrder.add(new ArrayList<>());
        }
        
        levelOrder.get(height).add(root.val);
        
        levelOrder(root.left, levelOrder, height + 1);
        levelOrder(root.right, levelOrder, height + 1);
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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