Binary Tree Level Order Traversal | Leetcode

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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

Confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

---

Solution: Maintain a list of list as the answer. Traverse in pre-order with the depth information. For each node of depth D, add it to the list with index D, where D=0 for the root node.

It does not matter if you traverse pre-order, in-order, or post-order. The result will be the same for any traversal order.

/**
 * 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>> answer = new ArrayList<>();
        traverse(answer, root, 0);
        return answer;
    }
    
    private void traverse(List<List<Integer>> answer, TreeNode node, int depth) {
        if (node == null) return;
        while (answer.size() <= depth) answer.add(new ArrayList<Integer>());
        answer.get(depth).add(node.val);
        traverse(answer, node.left, depth + 1);
        traverse(answer, node.right, depth + 1);
    }
}