【1错1对0】Binary Tree Level Order T

https://leetcode.com/problems/binary-tree-level-order-traversal-ii/

（来源：https://leetcode.com/problems/binary-tree-level-order-traversal-ii/）

关键：Binary Tree Level Order Traversal ，层级遍历。只要把每层的结果插在最前面，因此LindedList这个结构非常适合

1. 非递归

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> result = new LinkedList<>();
        
        if(root == null) {
            return result;
        }
        
        Queue<TreeNode> nodeQ = new LinkedList<>();
        nodeQ.offer(root);
        
        while(!nodeQ.isEmpty()) {
            int qLen = nodeQ.size();
            List<Integer> levelList = new ArrayList<>();
            for(int i=0; i<qLen; i++) {
                TreeNode node = nodeQ.poll();
                levelList.add(node.val);
                if(node.left != null) {
                    nodeQ.offer(node.left);
                }
                if(node.right != null) {
                    nodeQ.offer(node.right);
                }
            }
            
            result.add(0, levelList);
        }
        
        return result;
    }
    
}

2.递归

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> result = new LinkedList<>();
        
        if(root == null) {
            return result;
        }
        
        helper(root, result, 0);
        
        return result;
    }
    
    private void helper(TreeNode root, List<List<Integer>> result, int level) {
        if(root == null) {
            return;
        }
        
        if(level >= result.size()) {
            result.add(0, new LinkedList<>());
        }
        
        result.get(result.size()-1 - level).add(root.val);
        if(root.left != null) {
            helper(root.left, result, level+1);
        }
        if(root.right != null) {
            helper(root.right, result, level+1);
        }
        
    }
    
}