LeetCode（八）107. Binary Tree Level Order Traversal II

問題描述：
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]
]

求解思路：對比102. Binary Tree Level Order Traversal
這裏我再用一個棧來存儲所有的有序狀態下sublist，最後再從棧中pop出來，則形成逆序的list

代碼如下：

class Solution {
	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}
	
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> list = new LinkedList<List<Integer>>();
        Stack<List<Integer>> stack = new Stack<List<Integer>>();
        if(root == null) {
        	return list;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        
        while(!queue.isEmpty()) {
        	List<Integer> sublist = new LinkedList<Integer>();
        	int size = queue.size();
        	for(int i = 0; i < size; i++) {
        		TreeNode currentNode = queue.poll();
        		sublist.add(currentNode.val);
        		if(currentNode.left != null) {
        			queue.add(currentNode.left);
        		}
        		if(currentNode.right != null) {
        			queue.add(currentNode.right);
        		}
        	}
        	stack.push(sublist);
        }
        
        while(!stack.isEmpty()) {
        	list.add(stack.pop());
        }
        return list;
    }
    
}