Binary Tree Level Order Traversal II

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原题如下：
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).

这里写图片描述

我的算法：
1. 首先获得树的高度
2. 层序递归遍历二叉树(从左到右)

实现如下：

package leetCode;import java.util.ArrayList;import java.util.List;public class Solution {    //获得二叉树的高度，采用递归算法，简单实用    int depth(TreeNode root){        int res = 0;        if(root==null){            return 0;        }        int l = depth(root.left);        int r = depth(root.right);        res = Math.max(l, r)+1;        return res;    }    //当前层的高度为level，采用递归算法，简单实用，list为最终的结果list，    //list作参数（java对象作为参数），可以看做是引用传递，在方法中对list修改，能    //影响最终的list    void levelOrder(TreeNode root,int level,List<List<Integer>> list){        if(root==null){            return;        }        int toIndex = list.size()-1;        list.get(toIndex-level).add(root.val);        //从左到右递归层序递归遍历二叉树        levelOrder(root.left,level+1,list);        levelOrder(root.right,level+1,list);    }    public List<List<Integer>> levelOrderBottom(TreeNode root) {        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();        int dep = depth(root);        for(int i=0;i<dep;i++){            ArrayList<Integer> t = new ArrayList<Integer>();            res.add(t);        }        levelOrder(root,0,res);        return res;    }    public static void main(String[] args) {        TreeNode t = TreeNode.getDefaultTree();        List<List<Integer>> res = new ArrayList<List<Integer>>();        Solution s = new Solution();        res = s.levelOrderBottom(t);        for(int i=0;i<res.size();i++){            System.out.println(res.get(i));        }       }}

TreeNode的代码如下：

package leetCode;public class TreeNode {    int val;    TreeNode left;    TreeNode right;    TreeNode(int x) {        val = x;    }    //获得一颗默认的树，跟上面题中一样    public static TreeNode getDefaultTree(){        TreeNode t1 = new TreeNode(3);        TreeNode t2 = new TreeNode(9);        TreeNode t3 = new TreeNode(20);        TreeNode t4 = new TreeNode(15);        TreeNode t5 = new TreeNode(7);        t1.left = t2;        t1.right = t3;        t3.left = t4;        t3.right = t5;        return t1;    }}

原题地址：https://leetcode.com/problems/binary-tree-level-order-traversal-ii/

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