Leetcode -- 107. Binary Tree Level Order Traversal II

题目描述

题目难度：Easy

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).

AC代码1

DFS解法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<List<Integer>> resList = new ArrayList<>();
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        if(root == null) return resList;
        helper(root, 0);
        return resList;
    }
    
    private void helper(TreeNode root, int level){
        if(resList.size() == level)
            resList.add(0, new ArrayList<Integer>());
        if(root.left != null) helper(root.left, level + 1);
        if(root.right != null) helper(root.right, level + 1);
        resList.get(resList.size() - level - 1).add(root.val);
    }
}

AC代码2

BFS解法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> resList = new ArrayList<>();
        if(root == null) return resList;
        Queue<TreeNode> queue = new LinkedList<>();
        int start = 0, end = 1;
        queue.offer(root);
        TreeNode node;
        while(!queue.isEmpty()){
            start = 0;
            end = queue.size();
            List<Integer> list = new ArrayList<>();
            while(start < end){
                node = queue.poll();
                list.add(node.val);
                if(node.left != null) queue.offer(node.left);  
                if(node.right != null) queue.offer(node.right);
                start++;
            }
            resList.add(0, list);
        }
        return resList;
    }
}