Tree Algorithm: Find Closest Value In BST

算法刷题系列 Algorithm

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Difficulty：easy

Find Closest Value In BST

Write a function that takes in a Binary Search Tree (BST) and a target integer value and returns the closest value to that target value contained in the BST.
You can assume that there will only be one closest value.
Each BST node has an integer value , a left child node, and a right child node. A node is said to be a valid BST node if and only if it satisfies the BST property: its value is strictly greater than the values of every node to its left; its value is less than or equal to the values of every node to its right; and its children nodes are either valid BST nodes themselves or None / null .

input

{
     
  "tree": {
     
    "nodes": [
      {
     "id": "10", "left": "5", "right": "15", "value": 10},
      {
     "id": "15", "left": "13", "right": "22", "value": 15},
      {
     "id": "22", "left": null, "right": null, "value": 22},
      {
     "id": "13", "left": null, "right": "14", "value": 13},
      {
     "id": "14", "left": null, "right": null, "value": 14},
      {
     "id": "5", "left": "2", "right": "5-2", "value": 5},
      {
     "id": "5-2", "left": null, "right": null, "value": 5},
      {
     "id": "2", "left": "1", "right": null, "value": 2},
      {
     "id": "1", "left": null, "right": null, "value": 1}
    ],
    "root": "10"
  },
  "target": 12
}

output

13

Solution 1:
Average：O(log(N)) time | O(log(N)) space
Worst case：O(N) time | O(N) space

def findClosestValueInBst(tree, target):
    # Write your code here.
    return findClosestValueInBstHelper(tree, target, tree.value)

def findClosestValueInBstHelper(tree, target, closest):
	if tree is None:
		return closest
	if abs(target - closest) > abs(target - tree.value):
		closest = tree.value
	if target < tree.value:
		return findClosestValueInBstHelper(tree.left, target, closest)
	elif target > tree.value:
		return findClosestValueInBstHelper(tree.right, target, closest)
	else:
		return closest


# This is the class of the input tree.
class BST:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

Solution 2:
Average：O(log(N)) time | O(1) space
Worst case：O(N) time | O(1) space

def findClosestValueInBst(tree, target):
    # Write your code here.
    return findClosestValueInBstHelper(tree, target, tree.value)

def findClosestValueInBstHelper(tree, target, closest):
	currentNode = tree
	while currentNode is not None:
		if abs(target - closest) > abs(target - currentNode.value):
			closest = currentNode.value
		if target < currentNode.value:
			currentNode = currentNode.left
		elif target > currentNode.value:
			currentNode = currentNode.right
		else:
			break
	return closest

# This is the class of the input tree. 
class BST:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

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