【2错1对-1】Binary Tree Level Order Traversal II

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

日期	是否一次通过	comment
2019-01-21 13:23	否	思路太僵
2019-01-22 00:00	Y	理解分层遍历
2019-05-14 00:00	Y	歇菜

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（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> levelOrderBottom(TreeNode root) {
        List> result = new LinkedList<>();
        
        if(root == null) {
            return result;
        }
        
        Queue nodeQ = new LinkedList<>();
        nodeQ.offer(root);
        
        while(!nodeQ.isEmpty()) {
            int qLen = nodeQ.size();
            List levelList = new ArrayList<>();
            for(int i=0; i

2.递归

class Solution {
    public List> levelOrderBottom(TreeNode root) {
        List> result = new LinkedList<>();
        
        if(root == null) {
            return result;
        }
        
        helper(root, result, 0);
        
        return result;
    }
    
    private void helper(TreeNode root, List> 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);
        }
        
    }
    
}