Find closest value in BST

refer to : https://www.algoexpert.io/questions/Find%20Closest%20Value%20In%20BST

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Find Closest Value in BST

1. problem statement.

给定一个二叉搜索树和一个目标值，返回这个BST中跟目标值最接近的值closest value，假定只有一个最接近的值。

2. analysis

初始化一个closest value(root.value/Math.max_value)，从根部开始遍历二叉树，计算abs(curr - target) 和 abs(closest - target)，如果abs(curr - target) < abs(closest - target), 更新closest 为 curr. curr为当前节点的值。遍历的同时，比较target和curr的值，if target > curr, 直接去继续遍历curr所在的左子树，if target < curr, 直接去继续遍历curr所在的右子树，如果target == curr, 返回当前的closest。

3. recursive method

# average : O(logn) time | O(logn) space: recursive memory. (balanced BST)
# worst: O(n) time O(n) | space. (one branch BST)

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

def helper(tree, target, closest):
    if tree is None:
        return closest
    if abs(target-tree.value) < abs(target-closest):
        closest = tree.value
    if target > tree.value:
        return helper(tree.right,  target, closest)
    elif target < tree.value:
        return helper(tree.left, target, closest)
    else:
        return tree.value
        
# This is the class of the input tree. Do not edit.
class BST:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

import java.util.*;

class Program {
  public static int findClosestValueInBst(BST tree, int target) {
    // Write your code here.
    return helper(tree, target, tree.value);
  }
    
    public static int helper(BST tree, int target, int closest){
        if(Math.abs(target - tree.value) < Math.abs(target - closest)){
            closest = tree.value;
        }
        if(target < tree.value && tree.left != null){
            return helper(tree.left, target, closest);
        }else if(target > tree.value && tree.right != null){
            return helper(tree.right, target, closest);
      }else{
            return closest;
        }
    }    

  static class BST {
    public int value;
    public BST left;
    public BST right;

    public BST(int value) {
      this.value = value;
    }
  }
}

4. iterative method

# average : O(logn) time | O(1) space
# worst: O(n) time | O(1) space

省去重复调用helper function的步骤，用currNode追踪当前的节点，省空间。

# average : O(logn) time|O(1) space
# worst: O(n) time O(1)|space
def findClosestValueInBst(tree, target):
    # Write your code here.
    return helper(tree, target, tree.value)

def helper(tree, target, closest):
    currNode = tree
    while currNode is not  None:
        if abs(target-currNode.value) < abs(target-closest):
            closest = currNode.value
        if target > currNode.value:
            currNode = currNode.right
        elif target < currNode.value:
            currNode = currNode.left
        elif target == currNode.value:
            break    
    return closest
        
        
# This is the class of the input tree. Do not edit.
class BST:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

import java.util.*;

class Program {
  public static int findClosestValueInBst(BST tree, int target) {
    // Write your code here.
    return helper(tree, target, tree.value);
  }
    
    public static int helper(BST tree, int target, int closest){
        BST currNode = tree;
        while(currNode != null){
            if(Math.abs(target - currNode.value) < Math.abs(target - closest)){
                closest = currNode.value;
          }
          if(target < currNode.value){
              currNode = currNode.left;
          }else if(target > currNode.value){
              currNode = currNode.right;
        }else{
              break;
          }
        }
        return closest;
        
    }    

  static class BST {
    public int value;
    public BST left;
    public BST right;

    public BST(int value) {
      this.value = value;
    }
  }
}

some tips need to pay attention to.

BST currNode = tree; creat a new BST class and assign the value to tree.

while(currNode != null). don't forget this, and don't need to check currNode.left currNode.right is null or not in inner if statement

else{
             break;      if currNode.value = target, break

    }
outside the while loop, return closest.