Path Sum

Description

doc

Solutions

Recursive Version

/**
 * 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
 * @param {number} targetSum
 * @return {boolean}
 */
var hasPathSum = function (root, targetSum) {
  if (!root) {
    return false
  }

  const leftTarget = targetSum - root.val
  if (!leftTarget && !root.left && !root.right) {
    return true
  }

  return hasPathSum(root.left, leftTarget) || hasPathSum(root.right, leftTarget)
}

- Time Complexity: O(n)
- Space Complexity: O(1)

DFS

/**
 * 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
 * @param {number} targetSum
 * @return {boolean}
 */
var hasPathSum = function (root, targetSum) {
  if (!root) {
    return false
  }

  const queue = [root]
  while (queue.length) {
    const cur = queue.shift()

    const left = targetSum - cur.val
    if (!left && !cur.left && !cur.right) {
      return true
    }

    if (cur.left) {
      cur.left.val += cur.val // try to make it immutable in prod
      queue.push(cur.left)
    }

    if (cur.right) {
      cur.right.val += cur.val
      queue.push(cur.right)
    }
  }

  return false
}

- Time Complexity: O(n)
- Space Complexity: O(log(n)) ? maybe