lettcode-102:Binary Tree Level Order Traversal (Java)

Binary Tree Level Order Traversal

二叉树的层序遍历

两种方式：

1、用两个queue交替表示每一层的节点

2、用两个node，一个表示当前层的最后一个节点，一个表示下一层的最后一个节点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.Queue;
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {

        List<List<Integer>> list = new ArrayList<List<Integer>>();
        if(root == null)
            return list;
        Queue<TreeNode> queue = new ArrayDeque<TreeNode>();
        Queue<TreeNode> queue2 = new ArrayDeque<TreeNode>();
        queue.offer(root);

        List<Integer> listLevel = new ArrayList<Integer>();

        TreeNode curNode = null;
        while( !queue.isEmpty() || (!queue2.isEmpty())) {
            while(!queue.isEmpty()){
                curNode = queue.poll();
                listLevel.add(curNode.val);
                if(curNode.left != null) queue2.offer(curNode.left);
                if(curNode.right != null) queue2.offer(curNode.right);

                if(queue.isEmpty()){
                     list.add(listLevel);
                     listLevel = new ArrayList<Integer>();
                }
            }

            while(!queue2.isEmpty()){
                 curNode = queue2.poll();
                listLevel.add(curNode.val);
                if(curNode.left != null) queue.offer(curNode.left);
                if(curNode.right != null) queue.offer(curNode.right);

                if(queue2.isEmpty()){
                     list.add(listLevel);
                     listLevel = new ArrayList<Integer>();
                }
            }

        }

        /** 两个node，
         * lastNode:   当前层的最后一个节点
         * lastInqueue:表示某层最后进入队列的节点
        TreeNode curNode = root;
        TreeNode lastNode = root;
        TreeNode lastInQueue = root;
        while(!queue.isEmpty()){
            curNode = queue.poll();
            listLevel.add(curNode.val);

            if(curNode.left != null){
                queue.offer(curNode.left);
                lastInQueue = curNode.left;
            }
            if(curNode.right != null){
                queue.offer(curNode.right);
                lastInQueue = curNode.right;

            }

            if(curNode == lastNode){
                list.add(listLevel);
                // get last node
                lastNode = lastInQueue;
                listLevel = new ArrayList<Integer>();

            }
        }
        */

        return list;
    }
}