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Convert Binary Search Tree to Singly Linked List in C++
Suppose we have a binary tree; we have to convert this into a singly linked list (in place).
So, if the input is like
then the output will be
To solve this, we will follow these steps:
ser prev := null
Define a recursive function solve(), that will take root as input.
if root is null, then return
solve(right of root)
solve(left of root)
right of root := prev, left of root := null
prev := root
Let us see the following implementation to get better understanding:
Example
#include <bits/stdc++.h> using namespace std; class TreeNode{ public: int val; TreeNode *left, *right; TreeNode(int data){ val = data; left = NULL; right = NULL; } }; void insert(TreeNode **root, int val){ queue<TreeNode*> q; q.push(*root); while(q.size()){ TreeNode *temp = q.front(); q.pop(); if(!temp->left){ if(val != NULL) temp->left = new TreeNode(val); else temp->left = new TreeNode(0); return; }else{ q.push(temp->left); } if(!temp->right){ if(val != NULL) temp->right = new TreeNode(val); else temp->right = new TreeNode(0); return; }else{ q.push(temp->right); } } } TreeNode *make_tree(vector<int> v){ TreeNode *root = new TreeNode(v[0]); for(int i = 1; i<v.size(); i++){ insert(&root, v[i]); } return root; } class Solution { public: TreeNode* prev = NULL; void flatten(TreeNode* root) { if(!root) return; flatten(root->right); flatten(root->left); root->right = prev; root->left = NULL; prev = root; } }; main(){ vector<int> v = {1,2,5,3,4}; TreeNode *root = make_tree(v); Solution ob; (ob.flatten(root)); TreeNode *ptr = root; while(ptr != NULL && ptr->val != 0){ cout << ptr->val << ", "; ptr = ptr->right; } }
Input
{1,2,5,3,4}
Output
1, 2, 3, 4, 5,
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