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],
]
解法一:
每次将list保存到结果list的0下标的位置
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
LinkedList<TreeNode> que = new LinkedList<TreeNode>();
if(root==null) return res;
que.add(root);
while(!que.isEmpty()){
int size = que.size();
ArrayList<Integer> list = new ArrayList<Integer>();
for(int i=0;i<size;i++){
TreeNode cur = que.pop();
list.add(cur.val);
if(cur.left!=null){
que.add(cur.left);
}
if(cur.right!=null){
que.add(cur.right);
}
}
res.add(0,list);
}
return res;
}
}
解法二:
用递归实现层序遍历 与正常遍历不同的是,
先进行下一层递归,再把当前层的结果保存到res中
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
private ArrayList<ArrayList<Integer>> res;
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
res = new ArrayList<ArrayList<Integer>>();
LinkedList<TreeNode> que = new LinkedList<TreeNode>();
if(root==null) return res;
que.add(root);
gao(que);
return res;
}
public void gao(LinkedList<TreeNode> que){
int size = que.size();
if(size==0) return ;
ArrayList<Integer> list = new ArrayList<Integer>();
for(int i=0;i<size;i++){
TreeNode cur = que.pop();
list.add(cur.val);
if(cur.left!=null){
que.add(cur.left);
}
if(cur.right!=null){
que.add(cur.right);
}
}
gao(que);
res.add(list);
}
}