[Programming Problem] Flatten Binary Tree to Linked List

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.

[Problem Link]

Recursively run through the whole tree. During return of the recursion, root.right is set to left_linked_list (return from left recursion) and the end of left_linked_list.right is set to right_linked_list (return from right recursion). One gotcha, is the need to account for left_linked_list being empty/null. Also, don’t forget to set root.left to null.

 * 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 lastEl = function(root) {
    let curr = root;
    while (curr.right !== null) curr = curr.right;
    return curr;
var flatten = function(root) {
    if (!root) return null;
    let leftLL = flatten(root.left);
    let rightLL = flatten(root.right);
    root.left = null;
    //root.right = null;
    if (leftLL) {
        root.right = leftLL;
        lastEl(leftLL).right = rightLL;
    } else {
        root.right = rightLL;
    return root;

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