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]
]

解題思路：

首先想到的是二叉樹的層序遍歷，此題有兩個關注點，首先是如何用兩個計數變量來確定二叉樹每一層上子結點的個數，其次是Arraylist.add(0,templist)方法來實現倒序

算法如下：

/**
 * 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>> resultlist = new ArrayList<>();
        if(root==null)
            return resultlist;
        List<Integer> templist = new ArrayList<Integer>();
        LinkedList<TreeNode> s = new LinkedList<TreeNode>();
        s.add(root);
        int curIndex = 1;
        int nextIndex = 0;
        while(!s.isEmpty()){
            TreeNode node = s.poll();
            curIndex--;
            templist.add(node.val);
            if(node.left!=null){
                s.add(node.left);
                nextIndex++;
            }
            if(node.right!=null){
                s.add(node.right);
                nextIndex++;
            }
            if(curIndex==0){
                resultlist.add(0,templist);
                templist = new ArrayList<Integer>();
                curIndex = nextIndex ;
                nextIndex = 0;
            }
        }
        return resultlist;

    }
}