114. Flatten Binary Tree to Linked List

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114. Flatten Binary Tree to Linked List

Description

Difficulty: Medium

Related Topics: Linked List, Stack, Tree, Depth-First Search, Binary Tree

Given the root of a binary tree, flatten the tree into a “linked list”:

  • The “linked list” should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
  • The “linked list” should be in the same order as a pre-order traversal of the binary tree.

Example 1:

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Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

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Input: root = []
Output: []

Example 3:

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Input: root = [0]
Output: [0]

Constraints:

  • The number of nodes in the tree is in the range [0, 2000].
  • -100 <= Node.val <= 100

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

Hints/Notes

  • Recursive, think about if the left and right node has been flattened

Solution

Language: C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* curr = new TreeNode(-1);

    void flatten(TreeNode* root) {
        if (root == nullptr) {
            return;
        }

        flatten(root->left);
        flatten(root->right);

        TreeNode* left = root->left;
        TreeNode* right = root->right;

        root->left = nullptr;
        root->right = left;
        
        while (root->right != nullptr) {
            root = root->right;
        }
        root->right = right;
    }

};