zoukankan      html  css  js  c++  java
  • 156. Binary Tree Upside Down

    https://leetcode.com/problems/binary-tree-upside-down/description/

    Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

    For example:
    Given a binary tree {1,2,3,4,5},

        1
       / 
      2   3
     / 
    4   5
    

    return the root of the binary tree [4,5,2,#,#,3,1].

       4
      / 
     5   2
        / 
       3   1  
    

     

     

    Sol 1:

     

     

    This is not a very intuitive problem for me, I have to spent quite a while drawing figures to understand it. As shown in the figure, 1 shows the original tree, you can think about it as a comb, with 1, 2, 4 form the bone, and 3, 5 as the teeth. All we need to do is flip the teeth direction as shown in figure 2. We will remove the link 1--3, 2--5, and add link 2--3, and 4--5. And node 4 will be the new root.

    As the recursive solution, we will keep recurse on the left child and once we are are done, we found the newRoot, which is 4 for this case. At this point, we will need to set the new children for node 2, basically the new left node is 3, and right node is 1. Here is the recursive solution:

     

     

     

     

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    
    // recursion
    
    public class Solution {
        public TreeNode upsideDownBinaryTree(TreeNode root) {
            if (root == null || root.left == null){
                return root;
            }
            
            TreeNode newRoot = upsideDownBinaryTree(root.left);
            root.left.left = root.right; // node 2 left children
            root.left.right = root; //node 2 right children
            root.left = null;
            root.right = null;
            return newRoot;
            
        
            
        }
    }

    Sol 2:

    For the iterative solution, it follows the same thought, the only thing we need to pay attention to is to save the node information that will be overwritten.

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    
    // For the iterative solution, it follows the same thought, the only thing we need to pay attention to is to save the node information that will be overwritten.
    
    public class Solution {
        public TreeNode upsideDownBinaryTree(TreeNode root) {
            TreeNode curr = root;
            TreeNode next = null;
            TreeNode temp = null;
            TreeNode prev = null;
            
            while (curr != null){
                next = curr.left;
                
                // swapping nodes now, need temp to keep the previous right child
                // create the new left connection
                curr.left = temp;
                // create the new right connection, but store the previous one
                temp = curr.right;
                curr.right = prev;
                
                // advance
                prev = curr;
                curr = next;
            }
            
            return prev;
            
        
            
        }
    }
  • 相关阅读:
    人工智能第三课:数据科学中的Python
    人工智能第二课:认知服务和机器人框架探秘
    人工智能第一课:使用分类算法预测糖尿病
    如何加入Microsoft Teams 技术社区
    Python在Office 365 开发中的应用
    《Office 365开发入门指南》上市说明和读者服务
    基于Microsoft Graph打造自己的Timeline应用
    Office 365 应用开发的 .NET Core 模板库
    拥抱开源,Office 365开发迎来新时代
    Excel as a Service —— Excel 开发居然可以这么玩
  • 原文地址:https://www.cnblogs.com/prmlab/p/7294356.html
Copyright © 2011-2022 走看看