1028. Recover a Tree From Preorder Traversal

Description

We run a preorder depth-first search (DFS) on the root of a binary tree.

At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node.  If the depth of a node is D, the depth of its immediate child is D + 1.  The depth of the root node is 0.

If a node has only one child, that child is guaranteed to be the left child.

Given the output traversal of this traversal, recover the tree and return its root.

 

Example 1:

Input: traversal = "1-2--3--4-5--6--7"
Output: [1,2,5,3,4,6,7]

Example 2:

Input: traversal = "1-2--3---4-5--6---7"
Output: [1,2,5,3,null,6,null,4,null,7]

Example 3:

Input: traversal = "1-401--349---90--88"
Output: [1,401,null,349,88,90]

 

Constraints:

• The number of nodes in the original tree is in the range [1, 1000].
• 1 <= Node.val <= 109

 

Solutions

Solution: Depth-First Search

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * 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 {string} traversal
 * @return {TreeNode}
 */
const recoverFromPreorder = function (traversal) {
  const n = traversal.length;
  let index = 0;

  const generateTree = depth => {
    let level = 0;

    while (index < n && traversal[index] === '-') {
      level += 1;
      index += 1;
    }
    if (level !== depth) {
      index -= level;
      return null;
    }
    let value = '';

    while (index < n && traversal[index] !== '-') {
      value += traversal[index];
      index += 1;
    }
    const node = new TreeNode(+value);

    node.left = generateTree(depth + 1);
    node.right = generateTree(depth + 1);

    return node;
  };

  return generateTree(0);
};