标签：结果 题目 amp tree color 标记 lint The 对比
Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.
Example
Given root = {1}, target = 4.428571, return 1.
Notice
* Given target value is a floating point.
* You are guaranteed to have only one unique value in the BST that is closest to the target.
O(h)。注意不是O(logn)，没说平衡。
DFS树的递归。
夹逼法：
注意题目给的一个条件，是BST，所以要利用好这个条件。利用的方法就是分别找和lower bound和upper bound 。
1.lower bound: 比target只小一点点的数。如果根>=target了，那肯定得去左树找；如果根<target，那结果肯定是根或者右树里更好的一个。
2.upper bound: 对称同理。
3.综合，选两bound间更小的那个。
三者擂台法：
对比root，左边王者，右边王者。
这种方法没利用上BST信息进行剪枝，夹逼法每次砍半，这个每次两边还是都得去，遍历全树。但是此方法更general，什么树都可以这样做。
细节：
1.返回TreeNode不返回值，这样正好可以用返回null标记遇到null。
1.夹逼法
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: the given BST * @param target: the given target * @return: the value in the BST that is closest to the target */ public int closestValue(TreeNode root, double target) { // write your code here TreeNode lb = lowerBound(root, target); TreeNode ub = upperBound(root, target); int result = -1; double closestDist = Double.MAX_VALUE; if (lb != null && (target - lb.val) < closestDist) { closestDist = target - lb.val; result = lb.val; } if (ub != null && (ub.val - target) < closestDist) { closestDist = ub.val - target; result = ub.val; } return result; } private TreeNode lowerBound(TreeNode root, double target) { if (root == null) { return null; } if (root.val > target) { return lowerBound(root.left, target); } TreeNode right = lowerBound(root.right, target); if (right != null) { return right; } return root; } private TreeNode upperBound(TreeNode root, double target) { if (root == null) { return null; } if (root.val < target) { return upperBound(root.right, target); } TreeNode left = upperBound(root.left, target); if (left != null) { return left; } return root; } }
2.三者擂台法
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: the given BST * @param target: the given target * @return: the value in the BST that is closest to the target */ public int closestValue(TreeNode root, double target) { // write your code here if (root== null) { return -1; } double minDist = Math.abs(target - root.val); int minVal = root.val; if (root.left != null) { int leftClosest = closestValue(root.left, target); if (Math.abs(target - leftClosest) < minDist) { minDist = Math.abs(target - leftClosest); minVal = leftClosest; } } if (root.right != null) { int rightClosest = closestValue(root.right, target); if (Math.abs(target - rightClosest) < minDist) { minDist = Math.abs(target - rightClosest); minVal = rightClosest; } } return minVal; } }
lintcode900 - Closest Binary Search Tree Value - easy
标签：结果 题目 amp tree color 标记 lint The 对比
原文地址：https://www.cnblogs.com/jasminemzy/p/9565269.html
