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LeetCode-Binary Tree Level Order Traversal II

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]
]
/**
 * 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>> res=new ArrayList<List<Integer>>();
        ArrayList<TreeNode> pre=new ArrayList<TreeNode>();
        if(root==null){
            return res;
        }
        pre.add(root);
        List<Integer> rootList=new ArrayList<Integer>();
        rootList.add(root.val);
        res.add(rootList);
        while(!pre.isEmpty()){
             ArrayList<TreeNode> temp=new ArrayList<TreeNode>();
             List<Integer> tempValue=new ArrayList<Integer>();
             for(TreeNode node :pre){
                if(node.left != null){temp.add(node.left); tempValue.add(node.left.val);}
                if(node.right != null){temp.add(node.right); tempValue.add(node.right.val);}
             }
             if(!tempValue.isEmpty()){
                res.add(0, tempValue);
             }
             
             pre=new ArrayList<TreeNode>(temp);
        }
        return res;
    }
}

 

posted on 2016-05-10 05:05  IncredibleThings  阅读(107)  评论(0编辑  收藏  举报