First, we use recursive way. Successor 1publicclassSolution {2publicTreeNode inorderSuccessor(TreeNode root, TreeNode p) {3if(root ==null) {4returnnull;5}6if(root.val <=p.val) {7returninorderSuccessor(root.right, p);8}else{9TreeNode left =inorderSuccessor(root.left, p);10returnleft...
Find下一个Node in BST 给定一个node,找inorder traversal 里下一个。 最直接的想法是inorder遍历整个bst,然后在遍历的结果中查找。但是这样耗费多余的空间和时间。 geeksforgeek (1)有parent指针 在cc150上有很好的讲解 三种情况: i 当前node的right != null, 则返回leftmost of right subtree ii 当前node.r...
示例7: delete_element_from_BST ▲点赞 1▼ /* delete_element_from_BST: Searches the BST for the target(student_id), if it * is found, then remove it from the BST. Returns the new BST back to the caller. * Params: the BST to delete an item from and target to search the BST fo...
4.6 Write an algorithm to find the'next'node (i.e., in-order successor) of a given node in a binary search tree. You may assume that each node has a link to its parent. 这道题实际上考察的是一种叫线索二叉树的数据结构,而构造这种树的方法称之为Morris遍历法,在我之前的博客Binary Tree ...
}voidBSTNode::PrintTree() { PrintTreeInOrder(root); cout<<"\nFinished in PrintTree()"<<endl; } BSTNode *BSTNode::Search(BSTNode *node, int targetValue) { //The given target value is not found in BST if (node == NULL)
Note: There will be exactly one celebrity if he/she is in the party. Return the celebrity's label if there is a celebrity in the party. If there is no celebrity, return-1. 分析: 名流的定义竟然是其他所有人都知道他,而他却不知道其他任何人!好吧,承认也差不多是这么回事。言归正传,根据定...
因为有且仅有一个celebrity,而且所有人都知道celebrity。所以如果有celebrity,通过这层循环一定可以找到它。 第二层验证candidate是否符合celebrity条件。 1/*The knows API is defined in the parent class Relation.2boolean knows(int a, int b);*/34publicclassSolutionextendsRelation {5publicintfindCelebrity(int...