GuilinDev

Lc0107

05 August 2008

107 Binary Tree Level Order Traversal II

原题概述

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 

1

[3,9,20,null,null,15,7]
,

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5

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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[
  [15,7],
  [9,20],
  [3]
]

题意和分析

跟102比，DFS - 递归到最后一层再加,递的过程中先在总list的头部建好各层的list，最后取层数的时候从最后开始取，results.get(results.size() - 1 - level).add(current.val)； BFS - 加每一层list的时候直接加在总list 0的位置；

代码

DFS

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/**
 * 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>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (root == null) {
            return result;
        }
        levelOrderHelper(result, root, 0);
        return result;
    }
    private void levelOrderHelper(List<List<Integer>> result, TreeNode current, int level) {
        if (current == null) {
            return;
        }
        if (result.size() == level) {//先DFS一条路走到底把每一层的arraylist创建好，然后回溯的时候在call stack中取到level的值，把对应的层的值传进去
            result.add(new ArrayList<Integer>());
        }
        result.get(result.size() - 1 - level).add(current.val);

        levelOrderHelper(result, current.left, level + 1);
        levelOrderHelper(result, current.right, level + 1);
    }
}

BFS

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/**
 * 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>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> oneLevel = new ArrayList<>();
            while (size > 0) {
                TreeNode node = queue.poll();
                oneLevel.add(node.val);
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
                size--;
            }
            result.add(0, oneLevel);//唯一的不同点
        }
        return result;
    }
}
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