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Flatten Binary Tree to Linked List

Problem statement​

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:

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

Example 2:

Input: root = []Output: []

Example 3:

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)?

My solution​

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {void} Do not return anything, modify root in-place instead.
 */
var flatten = function(root) {
    let head = null;
    
    
    function dfs(node) {
        // console.log(node)
        if (!node) {
            return null;
        }
        let right = node.right;
        let left = node.left;        
        
            if (!head) {
            head = node
        } else if (head) {
            head.left = null;
            head.right = node;
            head = head.right;
        }
        
        dfs(left)
        dfs(right)
    }
    
    dfs(root)
    // console.log("head", head)
    // console.log("cacheHead", cacheHead)
    return head;
};