102. Binary Tree Level Order Traversal

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]
]

Code

/**
 * 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) {
        ArrayList result = new ArrayList();
        
        if (root == null) {
            return result;
        }
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        
        while(!queue.isEmpty()) {
            ArrayList<Integer> level = new ArrayList<Integer>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode head = queue.poll();
                level.add(head.val);
                if (head.left != null) {
                    queue.offer(head.left);
                }
                if (head.right != null) {
                    queue.offer(head.right);
                }          
            }
            result.add(level);
        }
        return result;
    }
}

解题思路

• 因为队列是先进后出的数据结构，使用队列queue来存储每一层结点；
• 使用while(!queue.isEmpty())控制总的结点个数；
• 每一层使用for(int i; i < size; i++)来控制单层的输出；
• 在每一层中使用level链表记录单层的输出。