![Java [Leetcode 107]Binary Tree Level Order Traversal II_小戴BOTAOY演示博客](/style/defalut/img/logo-bold.png){kind=logo}
题目描述：
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).文章来源地址:https://www.yii666.com/article/756105.html
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
解题思路：文章地址https://www.yii666.com/article/756105.html
运用广度优先搜索方法，运用队列的方法，每次遍历一层的叶节点，并把下一层的叶节点加入到队列中。每次遍历一层节点结束时候，将该层节点组成的list放到整体的list之前，实现由底到高的排列。
代码如下：
/**
* 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>> levelOrderBottom(TreeNode root) {
List<List<Integer>> list = new LinkedList<List<Integer>>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
if(root == null)
return list;
queue.offer(root);
while(!queue.isEmpty()){
int num = queue.size();
List<Integer> levelList = new LinkedList<Integer>();
for(int i = 0; i < num; i++){
if(queue.peek().left != null)
queue.offer(queue.peek().left);
if(queue.peek().right != null)
queue.offer(queue.peek().right);
levelList.add(queue.poll().val);
}
list.add(0, levelList);
}
return list;
}
}
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