Medium 107题 Binary Tree Level Order Traversal II

Question:

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]]

Solution:

booleanadd(E e)

Appends the specified element to the end of this list.

voidadd(int index, E element)

Inserts the specified element at the specified position in this list.

/** * 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>> ans =new LinkedList<List<Integer>>();        Queue<TreeNode> q=new LinkedList<TreeNode>();        if(root == null) return ans;        q.offer(root);        while(!q.isEmpty())        {          //  TreeNode tmp=q.peek();            int levelNum = q.size();            List<Integer> curlevel=new LinkedList<Integer>();            for(int i=1;i<=levelNum;i++){                if(q.peek().left!=null){                    q.offer(q.peek().left);                //     curlevel.add(q.peek().left.val); //不可以在这里就入curlevel,因为这样无法打印出root.                }                if(q.peek().right!=null){                    q.offer(q.peek().right);               //     curlevel.add(q.peek().right.val);                }                curlevel.add(q.poll().val);            }            ans.add(0,curlevel);        }        return ans;         }}

DFS

public List<List<Integer>> levelOrderBottom(TreeNode root) {          List<List<Integer>> ans =new LinkedList<List<Integer>>();          f(ans,root,0);          return ans;     }     public void f(List<List<Integer>> ans, TreeNode root, int level){         if(root==null) return;         if(level>=ans.size())         {             ans.add(0,new LinkedList<Integer>());         }         f(ans,root.left,level+1);         f(ans,root.right,level+1);         ans.get(ans.size()-level-1).add(root.val); //ans.size()-level-1 //从第一个列表中开始加，每回溯一层因为level--，所以就相当于于往下一个列表中加     }