Leet Code OJ 107. Binary Tree Level Order Traversal II [Difficulty: Easy]

Posted 2020-06-24 Lnho

题目：
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,#,#,15,7},

return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]

翻译：
给定一个二叉树，返回它的节点的从底部到头部的层序遍历结果。

代码：

/**
 * 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) {
        List<List<Integer>> result=levelOrder(root);
        Collections.reverse(result);
        return result;
    }
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result=new ArrayList<>();
        if(root==null){
            return result;
        }
        List<Integer> one=new ArrayList<>();
        one.add(root.val);
        result.add(one);
        List<List<Integer>> left=levelOrder(root.left);
        List<List<Integer>> right=levelOrder(root.right);
        for(int i=0;i<left.size()||i<right.size();i++){
            List<Integer> item=new ArrayList<>();
            if(i<left.size()){
                for(Integer k:left.get(i)){
                    item.add(k);
                }
            }
            if(i<right.size()){
                for(Integer k:right.get(i)){
                    item.add(k);
                }
            }
            result.add(item);
        }
        return result;
    }
}