Binary Tree Level Order Traversal

Problem

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],

return its level order traversal as:

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

Solution (Java)

/**
 * Definition for a binary tree node.
 * 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;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        queue.add(null);
        List<Integer> level = new ArrayList<>();
        while (!queue.isEmpty()) {
            root = queue.remove();
            while (!queue.isEmpty() && root != null) {
                level.add(root.val);
                if (root.left != null) {
                    queue.add(root.left);
                }
                if (root.right != null) {
                    queue.add(root.right);
                }
                root = queue.remove();
            }
            result.add(level);
            level = new ArrayList<>();
            if (!queue.isEmpty()) {
                queue.add(null);
            }
        }
        return result;
    }
}

Solution 1

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function(root) {
  if (!root) return [];
  return helper([[root]], 0);
};

var helper = function (res, level) {
  var now = res[level];
  var next = [];

  for (var i = 0; i < now.length; i++) {
    if (now[i].left) next.push(now[i].left);
    if (now[i].right) next.push(now[i].right);
    now[i] = now[i].val;
  }

  if (next.length) {
    res.push(next);
    helper(res, level + 1);
  }

  return res;
};

Explain:

bfs

Complexity:

Time complexity : O(n).
Space complexity : O(n).

Solution 2

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function(root) {
  return helper([], root, 0);
};

var helper = function (res, root, level) {
  if (root) {
    if (!res[level]) res[level] = [];
    res[level].push(root.val);
    helper(res, root.left, level + 1);
    helper(res, root.right, level + 1);
  }
  return res;
};

Explain:

dfs

Complexity:

Time complexity : O(n).
Space complexity : O(n).