102. 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],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

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).