zoukankan      html  css  js  c++  java
  • [LC] 107. Binary Tree Level Order Traversal II

    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,null,null,15,7],

        3
       / 
      9  20
        /  
       15   7
    

    return its bottom-up level order traversal as:

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

    Solution 1:
    BFS
    /**
     * 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>> arrList = new ArrayList<>();
            if (root == null) {
                return arrList;
            }
            Queue<TreeNode> queue = new LinkedList<>();
            queue.offer(root);
            while (!queue.isEmpty()) {
                int size = queue.size();
                List<Integer> list = new ArrayList<>();
                for (int i = 0; i < size; i++) {
                    TreeNode cur = queue.poll();
                    list.add(cur.val);
                    if (cur.left != null) {
                        queue.offer(cur.left);
                    }
                    if (cur.right != null) {
                        queue.offer(cur.right);
                    }
                }
                arrList.add(0, list);
            }
            return arrList;
        }
    }

    Solution 2:

    /**
     * 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>> arrList = new ArrayList<>();
            helper(arrList, 0, root);
            return arrList;
        }
        
        private void helper(List<List<Integer>> res, int depth, TreeNode root) {
            if (root == null) {
                return;
            }
            // backtrack case, resSize >= depth
            if (depth >= res.size()) {
                res.add(0, new LinkedList<>());
            }
            res.get(res.size() - 1 - depth).add(root.val);
            helper(res, depth + 1, root.left);
            helper(res, depth + 1, root.right);
        }
    }
  • 相关阅读:
    day5
    \_\_setitem\_\_和\_\_getitem和\_\_delitem__
    描述符(\_\_get\_\_和\_\_set\_\_和\_\_delete\_\_)
    \_\_getattribute\_\_
    面向对象进阶小结
    property装饰器
    super()方法详解
    菱形继承问题
    类的派生
    类的继承
  • 原文地址:https://www.cnblogs.com/xuanlu/p/12151471.html
Copyright © 2011-2022 走看看