Shawngbk / Leecode

Questions of Leecode
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102. Binary Tree Level Order Traversal #32

Open Shawngbk opened 7 years ago

Shawngbk commented 7 years ago

/**

public class Solution { public List<List> levelOrder(TreeNode root) { List<List> list = new LinkedList<List>(); if(root == null) return list; //创建一个队列,用于存放所有节点 Queue curLevel = new LinkedList(); curLevel.add(root); while(!curLevel.isEmpty()) { //创建一个List记录当前层所有节点值 int size = curLevel.size(); List curList = new LinkedList(); for(int i = 0; i<size; i++) { //poll()获取并移除此列表的头结点 TreeNode curNode = curLevel.poll(); curList.add(curNode.val); //记录每一层的节点值到List中 if(curNode.left != null) { curLevel.offer(curNode.left); } if(curNode.right != null) { curLevel.offer(curNode.right); } } //将当前List添加到List<List>中 list.add(curList); } return list; } }

Shawngbk commented 7 years ago

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