107. 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,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its bottom-up level order traversal as:
[
  [15,7],
  [9,20],
  [3]
]

難度：easy

題目：給定二叉樹，返回由底到上的層次遍歷值。

思路：層次遍歷

Runtime: 1 ms, faster than 97.69% of Java online submissions for Binary Tree Level Order Traversal II.
Memory Usage: 26.4 MB, less than 80.95% of Java online submissions for Binary Tree Level Order Traversal II.

/**
 * 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) {
        LinkedList<List<Integer>> result = new LinkedList<>();
        if (null == root) {
            return result;
        }
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        int levelCnt = 1;
        while (!queue.isEmpty()) {
            int cnt = 0;
            List<Integer> llt = new ArrayList<>();
            for (int i = 0; i < levelCnt; i++) {
                TreeNode node = queue.poll();
                llt.add(node.val);
                if (node.left != null) {
                    queue.add(node.left);
                    cnt++;
                }
                if (node.right != null) {
                    queue.add(node.right);
                    cnt++;
                }
            }
            levelCnt = cnt;
            result.addFirst(llt);
        }
        
        return result;
    }
}