Binary Tree Level Order Traversal

ID: 69; easy

LintCode 炼码

Solution 1 (Java)

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: A Tree
     * @return: Level order a list of lists of integer
     */
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();
        if (root == null) return list;

        Queue<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode curr = q.poll();
                level.add(curr.val);
                if (curr.left != null)
                    q.offer(curr.left);
                if (curr.right != null)
                    q.offer(curr.right);
            }
            list.add(level);
        }
        return list;
    }
}

Notes

• Classic BFS