Path Sum II

Description

doc

Solutions

First Idea

/**
 * 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 {number[][]}
 */
var pathSum = function (root, targetSum) {
  if (root === null) {
    return []
  }

  const leftVal = targetSum - root.val
  if (root.left === null && root.right === null) {
    return leftVal ? [] : [[root.val]]
  }

  const left = pathSum(root.left, leftVal)
  const right = pathSum(root.right, leftVal)

  return [...left, ...right].map((v) => [root.val, ...v])
}

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

BFS

/**
 * 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 {number[][]}
 */
var pathSum = function (root, targetSum) {
  if (root === null) {
    return []
  }

  root.vals = [root.val]
  let queue = [root]
  let result = []

  while (queue.length) {
    const cur = queue.shift()

    if (cur.left === null && cur.right === null && cur.val === targetSum) {
      result.push(cur.vals)
      continue
    }

    if (cur.left) {
      cur.left.vals = [...cur.vals, cur.left.val]
      cur.left.val += cur.val
      queue.push(cur.left)
    }

    if (cur.right) {
      cur.right.vals = [...cur.vals, cur.right.val]
      cur.right.val += cur.val
      queue.push(cur.right)
    }
  }

  return result
}

• Time Complexity: O(n)
• Space Complexity: O(nlogn)