Binary Tree Level Order Traversal II

题目描写叙述

3
/ \
9  20
/  \
15   7

[
[15,7],
[9,20],
[3]
]

解题思路

具体代码

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
//Binary Tree Level Order Traversal
class TreeNodeWithDepth
{
TreeNode treeNode;
int depth;

public TreeNodeWithDepth(TreeNode treeNode, int depth)
{
this.treeNode = treeNode;
this.depth = depth;
}
}

public class Solution {
//Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
//二叉树的层次遍历 2 最底层最先遍历
public List<List<Integer>> levelOrderBottom(TreeNode root) {
int depth = 1;
int lastLevel = 1; //while中用来标示上一次节点的level
List<List<Integer>> resultsList = new LinkedList<List<Integer>>();  //注意初始化 List是接口类 //这点和levelOrder不同使用LinkedList
Queue<TreeNodeWithDepth> queue = new LinkedList<>();
TreeNodeWithDepth pNode = null;
if(root != null)
{
pNode = new TreeNodeWithDepth(root, 1);
queue.offer(pNode);
}
List<Integer> listPerLevel = new ArrayList<>();//每层的list
while(!queue.isEmpty())
{
pNode = queue.poll();//程序增加时已经推断不为空节点

if(pNode.depth > lastLevel)
{
//说明是新的一层
if(listPerLevel.size()!=0)
{
//把上一层的列表增加result中 插入到第一个位置
resultsList.add(listPerLevel);
}
listPerLevel = new ArrayList<>(); //又一次另起一个listPerLevel
listPerLevel.add(pNode.treeNode.val);//在新列表中增加当前节点
lastLevel = pNode.depth;//更新lastLevel
}
else
{
listPerLevel.add(pNode.treeNode.val);
}
//确定子节点深度
depth = pNode.depth+1;

//子节点入队列
if(pNode.treeNode.left != null)
{
queue.offer(new TreeNodeWithDepth(pNode.treeNode.left, depth));
}
if(pNode.treeNode.right != null)
{
queue.offer(new TreeNodeWithDepth(pNode.treeNode.right, depth));
}

}

//while循环中最后一次的listPerLevel也要增加
if(listPerLevel.size()>0)
{
resultsList.add(0,listPerLevel);
}
return resultsList;
}
}