LeetCode-Binary Tree Level Order Traversal II

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本文介绍了一种解决LeetCode上二叉树层次遍历II问题的方法，提供了两种Java实现方案，一种使用栈结构从根节点开始遍历，另一种采用层次遍历的方式并反转结果。

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作者：disappearedgod

文章出处：http://blog.youkuaiyun.com/disappearedgod/article/details/38896661

时间：2014-5-26

题目

Binary Tree Level Order Traversal II

Total Accepted: 16336 Total Submissions: 52120 My 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},

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;
    }
}