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).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
/ \
9 20
/ \
15 7

return its bottom-up level order traversal as:

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

这里要求层序遍历,而且结果是从下往上,不是从上往下。

思路:1.可以继续从上往下添加到list中,最后反过来放到另一个list中返回

    2:使用list中的add(index,element)方法,该方法是往index位置插入元素,该位置上原来的元素以及后面所有元素往后移,这里只需往0处一直插入就行,就会自动往后移。因为插入会移动后面所有元素,所以此时适合linkedList。

代码1:

class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> tem=new ArrayList<List<Integer>>();
if(root==null) return tem;
Queue<TreeNode> q=new LinkedList<>();
List<Integer> list=new ArrayList<>();
q.add(root);
while(!q.isEmpty()){
int size=q.size();
for(int i=0;i<size;i++){
TreeNode node=q.poll();
list.add(node.val);
if(node.left!=null) q.add(node.left);
if(node.right!=null) q.add(node.right);
}
tem.add(list); list=new ArrayList();
} List<List<Integer>> result=new ArrayList<List<Integer>>();
for(int i=tem.size()-1;i>=0;i--)
result.add(tem.get(i));
return result; }
}

代码2:

  public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> tem=new LinkedList<List<Integer>>();
if(root==null) return tem;
Queue<TreeNode> q=new LinkedList<>();
List<Integer> list=new ArrayList<>();
q.add(root);
while(!q.isEmpty()){
int size=q.size();
for(int i=0;i<size;i++){
TreeNode node=q.poll();
list.add(node.val);
if(node.left!=null) q.add(node.left);
if(node.right!=null) q.add(node.right);
}
tem.add(0,list); list=new ArrayList();
} return tem; }
												

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