importcom.kaesar.leetcode.TreeNode;importjava.util.LinkedList;importjava.util.Queue;publicclassLeetCode_222 {/*** 二叉树层序遍历** @param root* @return*/publicstaticintcountNodes(TreeNoderoot) {// 如果根节点为空，即为空树，说明没有节点，直接返回0if (root==null) {return; }// 利用队列...
1intcountNodes(TreeNode*root) {2inth = getHeight(root), count =0;3while(root) {4//情况一:左孩子是满二叉树(右孩子可能是完全二叉树,也可能是满二叉树)5if(getHeight(root -> right) == h -1) {6count +=1<< h;//根节点 + 左孩子7root = root ->right;8}9//情况二:左孩子是完全...
Given a complete binary tree, count the number of nodes. Definition of a complete binary tree from Wikipedia: In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h...
https://leetcode.com/problems/count-complete-tree-nodes/ 题目: complete Definition of a complete binary tree fromWikipedia: In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have betwee...
Given a complete binary tree, count the number of nodes. Note: Definition of a complete binary tree from Wikipedia: In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It ...
3.当右子树不存在而左子树存在时,递归1 + countNodes(root->left); 4.当左右子树都不存在时,递归结束,返回1。 代码 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL)...
//#222Description: Count Complete Tree Nodes | LeetCode OJ 解法1:可以二分。 // Solution 1: Binary search. 代码1 //Code 1 223 Rectangle Area // #223 矩形面积 描述:给定两个横平竖直的矩形,求总覆盖面积。 //#223Description: Rectangle Area | LeetCode OJ ...
题目链接: https://leetcode-cn.com/problems/count-complete-tree-nodes难度:中等 通过率:57.4% 题目描述:给出一个 完全二叉树 ,求出该树的节点个数。 说明: 完全二叉树的定义如下:在完全二叉树中,除了最…
classSolution{publicintcountNodes(TreeNode root){if(root==null){return0;}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=(high-low+1)/2+low;if(exists(root,level,mid)){low=mid;}els...
270.Closest-Binary-Search-Tree-Value (M+) 095.Unique-Binary-Search-Trees-II (H) 094.Binary Tree Inorder Traversal (H-) 110.Balanced-Binary-Tree (M+) 222.Count-Complete-Tree-Nodes (M+) 099.Recover-Binary-Search-Tree (H) 114.Flatten-Binary-Tree-to-Linked-List (M+) 098.Validate-Binar...