《原神攻略》淵下宮厄瑞玻斯的祕密系列任務攻略
阿新 • • 發佈:2022-01-08
一、題目
輸入:root = [1,2,3,4,5,6]
輸出:6
二、解法
一般的做法是用bfs或dfs遍歷節點,時間和空間複雜度是O(n)。
要利用完全二叉樹這個性質,首先求出樹的層數level(root是0層),然後二分查詢,判斷最高層節點是否存在。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public int countNodes(TreeNode root) { if(root==null) return 0; int level=0; TreeNode node=root; while(node.left!=null){ level++; node=node.left; } int low=1<<level,high=(1<<(level+1))-1; while(low<high){ int mid=low+(high-low+1)/2; if(exists(root,level,mid)){ low=mid; }else{ high=mid-1; } } return low; } boolean exists(TreeNode root,int level,int k){ int bits=1<<(level-1); TreeNode node=root; while(node!=null&&bits!=0){ if((k&bits)==0){ node=node.left; }else{ node=node.right; } bits>>=1; } return node!=null; } }