but given a node, how to check if it is or not, we should do this: isBalanced(root) = Math.abs(height(root.left) - height(root.right)) < 1 classSolution{publicbooleanisBalanced(TreeNode root){if(root ==null)returntrue;if(Math.abs(height(root.left) - height(root.right)) >1) {...
int GetDepth(TreeNode *root) {if(root == NULL)return0;returnmax(GetDepth(root->left), GetDepth(root->right)) +1; } bool GetAns(TreeNode *root, int dep) {//只有根节点的情况,不用判空,因为不会递归到那if(dep == depth) {returntrue; }if(dep < depth -1) {if(root->left == ...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode ...
Given a binary tree, determine if it is acomplete binary tree. 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 between 1 a...
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The tree will have between 1 and 100 nodes. 这道题给了一棵二叉树,让我们判断是否是一棵完全二叉树 Complete Binary Tree,通过题目中的解释可知,完全二叉树除了最后一行之外,所有位置都是满员的,而且最后一行的结点都是尽可能靠左的,注意跟完满二叉树 Full Bianry Tree 区分开来。最简单直接的方法就是按层序遍...
1classSolution2{3public:4boolisCompleteTree(TreeNode*root)5{6if(root ==NULL)7returnfalse;89queue<TreeNode *>q;10q.push(root);11boolmustHaveNoChild =false;12boolresult =true;13while(!q.empty())14{15TreeNode* pNode =q.front();16q.pop();17if(mustHaveNoChild)//如果一个节点没有子...
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 nodes inclusive at the last level h. Example 1:
Try explaining your solution to someone and see if it makes sense ot them. Don't code anything for this problem Solution Hopefully this problem sounds familiar! We can use a binary search to search for an intger since the list is already sorted! This means we can find the item inO(logn...