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900. Closest Binary Search Tree Value
https://www.lintcode.com/problem/closest-binary-search-tree-value/description?_from=ladder&&fromId=1
1. 非递归方法:求BST中跟target最近的数字。我们先设置一个min = root.val, 然后用iterative的方法尝试更新min,然后比较target与root的大小,进行二分查找。
public int closestValue(TreeNode root, double target) { // write your code here int min = root.val; while(root != null) { min = Math.abs(target - root.val) < Math.abs(target - min) ? root.val : min; root = root.val > target ? root.left : root.right; } return min; }
2. 递归
1. 比较target和root.val, => 求child是为了递归
2. if(child == null) return root.val;
3. 求 childClosest = closestValue(child, target)
4. 比较 root.val 和childClosest
public int closestValue(TreeNode root, double target) { // write your code here TreeNode child = root.val > target ? root.left : root.right; if(child == null) { return root.val; } int childClosest = closestValue(child, target); return Math.abs(root.val - target) > Math.abs(childClosest - target) ? childClosest : root.val; }
596. Minimum Subtree
https://www.lintcode.com/problem/minimum-subtree/description?_from=ladder&&fromId=1
public class Solution { /** * @param root: the root of binary tree * @return: the root of the minimum subtree */ public TreeNode findSubtree(TreeNode root) { // write your code here ResultType result = helper(root); return result.minSubtree; } public ResultType helper(TreeNode node) { if(node == null) { return new ResultType(null, Integer.MAX_VALUE, 0); } ResultType leftResult = helper(node.left); ResultType rightResult = helper(node.right); ResultType result = new ResultType( node, leftResult.sum + rightResult.sum + node.val, leftResult.sum + rightResult.sum + node.val ); if(leftResult.minSum <= result.minSum) { result.minSum = leftResult.minSum; result.minSubtree = leftResult.minSubtree; } if(rightResult.minSum <= result.minSum) { result.minSum = rightResult.minSum; result.minSubtree = rightResult.minSubtree; } return result; } }
5 - Binary Tree & Tree-based DFS
标签:nod https href ini mini ntc 数字 color bst
原文地址:https://www.cnblogs.com/jenna/p/10804263.html