zoukankan      html  css  js  c++  java
  • LeetCode Flatten Binary Tree to Linked List

    Given a binary tree, flatten it to a linked list in-place.
    
    For example,
    Given
    
             1
            / 
           2   5
          /    
         3   4   6
    The flattened tree should look like:
       1
        
         2
          
           3
            
             4
              
               5
                
                 6
    

    感觉以前写过blog了,先来个递归版本(注意left指针全部要设置成NULL)

    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    private:
        TreeNode* last;
    public:
        void flatten(TreeNode* root) {
            last = NULL;
            dfs(root);
        }
        
        void dfs(TreeNode* root) {
            if (root == NULL) {
                return;
            }
            
            TreeNode* r = root->right;
            TreeNode* l = root->left;
            
            if (last != NULL) {
                last->right = root;
            }
    
            last = root;
            last->left = NULL;
    
            dfs(l);
            dfs(r);
        }
    };
    

    非递归版本discuss里找的。

    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
        void flatten(TreeNode* root) {
            while (root) {
                if (root->left && root->right) {
                    TreeNode* largest = root->left;
                    while (largest->right) {
                        largest = largest->right;
                    }
                    largest->right = root->right;
                }
                if (root->left) {
                    root->right = root->left;
                }
                root->left = NULL;
                root = root->right;
            }
        }
    };
    
  • 相关阅读:
    4-1 R语言函数 lapply
    3-6 向量化操作
    3-5 处理缺失值
    3-4 列表的子集
    3-3 数据框的子集
    3-2 矩阵的子集
    bootstrap 模式对话框
    手机端 超链接 识别电话号码
    jQuery设置和获取HTML、文本和值
    TP 框架 ajax[利用异步提交表单]
  • 原文地址:https://www.cnblogs.com/lailailai/p/4611599.html
Copyright © 2011-2022 走看看