LeetCode 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:

BFS

/**
 * 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>> result = new ArrayList<List<Integer>>();
        if (root == null) return result;
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            List<Integer> row = new ArrayList<Integer>();
            while (size > 0) {
                TreeNode current = q.poll();
                row.add(current.val);
                if (current.left != null) q.offer(current.left);
                if (current.right != null) q.offer(current.right);
                size--;
            }
            result.add(row);
        }
        return result;
    }
}

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