Binary Tree Level Order Traversal2

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.

/**

 * Definition for binary tree

 * public class TreeNode {

 *     int val;

 *     TreeNode left;

 *     TreeNode right;

 *     TreeNode(int x) { val = x; }

 * }

 */

public class Solution {

    public List<List<Integer>> levelOrderBottom(TreeNode root) {

        

    }

}

分析：

给你一个二叉树，返回它的从底向上的节点值排序，（从左到右，从叶子到根）

public class BinaryTreeLevelOrderTraversal2{    public class TreeNode{        int val;        TreeNode left;        TreeNode right;        TreeNode(int x){val=x;}}    public List<List<Integer>>levelOrderBottom (TreeNode root){}    public bfs(){        }}

相对于该问题的版本1，其就是把结果反着输出

返回结果前加一句：

Collections.reverse(result); 

即可。

答案：

    public List<List<Integer>> levelOrder(TreeNode root) {    List<List<Integer>> result= new ArrayList<List<Integer>>();    if(root==null) return result;    Queue<TreeNode> queue = new LinkedList<TreeNode>();         queue.add(root);    int curLevelCount=1;    int nextLevelCount=0;    while(!queue.isEmpty()){    ArrayList<Integer> level = new ArrayList<Integer>();    while(curLevelCount>0){    TreeNode temp = queue.remove();    curLevelCount--;    if(temp.left!=null) {nextLevelCount++; queue.add(temp.left);}    if(temp.right!=null) {nextLevelCount++; queue.add(temp.right);}    level.add(temp.val);    }    curLevelCount=nextLevelCount;    nextLevelCount=0;    result.add(level);        }    Collections.reverse(result);    return result;    }