Binary Tree Level Order Traversal

Given a binary tree, return thelevel ordertraversal 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

Recursive solution.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        // each level is one index in result list (we pass level as parameter in the method, if it does not exist, we create it)
        traverse(root, result, 0);
        return result;
    }

    public void traverse(TreeNode node, List<List<Integer>> list, int level) {
        if (node == null) return;
        if (level == list.size()) {
            list.add(new ArrayList<>());
        }
        list.get(level).add(node.val);
        traverse(node.left, list, level + 1);
        traverse(node.right, list, level + 1);
    }
}

Non recursive solution.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ans = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        if (root != null) {
            q.offer(root);
        }
        TreeNode cur;
        while (!q.isEmpty()) {
            int size = q.size();
            List<Integer> subAns = new LinkedList<Integer>();
            for (int i = 0; i < size; ++i) {        // traverse nodes in the same level
                cur = q.poll();
                subAns.add(cur.val);                // visit the root
                if (cur.left != null) {
                    q.offer(cur.left);              // push left child to queue if it is not null
                }
                if (cur.right != null) {
                    q.offer(cur.right);             // push right child to queue if it is not null
                }
            }
            ans.add(subAns);
        }
        return ans;
    }
}