LeetCode-Binary Tree Level Order Traversal II

作者：disappearedgod

文章出处：http://blog.csdn.net/disappearedgod/article/details/38896661

时间：2014-5-26

题目

Binary Tree Level Order Traversal II

 Total Accepted: 16336 Total Submissions: 52120My Submissions

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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

解法

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public ArrayList<ArrayList<Integer>> levelOrderBottom1(TreeNode root) {        ArrayList<ArrayList<Integer>> retlist = new ArrayList<ArrayList<Integer>>();        ArrayList<Integer> list = new ArrayList<Integer>();        Stack<TreeNode> s= new Stack<TreeNode>();        if(root == null)            return retlist;        s.push(root);        list.add(root.val);        while(!s.isEmpty()){            //list.clear();            list = new ArrayList<Integer>();            TreeNode t = s.pop();            if(t!=null){                                if(t.left!=null)                {                        list.add(t.left.val);                    s.push(t.left);                }                if(t.right!=null){                    list.add(t.right.val);                    s.push(t.right);                }            }            retlist.add(list);        }        ArrayList<ArrayList<Integer>> ret2 = new ArrayList<ArrayList<Integer>>();          for(int i=retlist.size()-1; i>=0; i--){              ret2.add(retlist.get(i));          }                    return ret2;      }    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {    // Start typing your Java solution below    // DO NOT write main() function    ArrayList<ArrayList<Integer>> result=new ArrayList<ArrayList<Integer>>();        if(root==null) return result;    Stack<ArrayList<Integer>> stack=new Stack<ArrayList<Integer>>();    ArrayList<TreeNode> list=new ArrayList<TreeNode>();    list.add(root);    while(!list.isEmpty())    {        ArrayList<TreeNode> Tplist=new ArrayList<TreeNode>();        ArrayList<Integer> level=new ArrayList<Integer>();        while(!list.isEmpty())          {            TreeNode node=list.remove(0);            level.add(node.val);            if(node.left!=null) Tplist.add(node.left);            if(node.right!=null) Tplist.add(node.right);        }        stack.push(level);        list=Tplist;    }        while(!stack.isEmpty())        {            result.add(stack.pop());        }    return result;    }}

结果

返回

 LeetCode Solution(持续更新，java>c++)