LeetCode: Flatten Binary Tree To Linked List

Readme.md

Flatten Binary Tree to Linked List

http://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/

Problem Description

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

Structures

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

Solution-Recursive-1.cpp

/*
 * Time complexity: O(n) (where n is the number of nodes).
 * Space: O(1)
 */

class Solution {
public:

    TreeNode* flattenT(TreeNode *root) {
        if (!root || !root->left && !root->right) return root;

        TreeNode *node1 = flattenT(root->left),
                 *node2 = flattenT(root->right);
                 
        if (node1) node1->right = root->right;
        if (root->left) root->right = root->left; 
        root->left = NULL;
        
        return (node2) ? node2 : node1;
    }
    
    void flatten(TreeNode *root) {
        flattenT(root);
    }
};

How's this O(1) space complexity? On each recursive call you are storing 2 variables (node1 and node2) on the stack.

I guess it doesn't store 2 variables on each recursive call, it stores 3 variables, node1, node2 and root.