* @param root: The root of the binary search tree. * @param value: Remove the node with given value. * @return: The root of the binary search tree after removal. */ publicTreeNode removeNode(TreeNode root,intvalue) { // write your code here TreeNode dummy =newTreeNode(0); dummy....
1publicclassSolution {2/**3*@paramroot: The root of the binary search tree.4*@paramvalue: Remove the node with given value.5*@return: The root of the binary search tree after removal.6*/7publicTreeNode removeNode(TreeNode root,intvalue) {8if(root ==null) {9returnnull;10}1112if(ro...
Given a root of Binary Search Tree with unique value for each node. Remove the node with given value. If there is no such a node with given value in the binary search tree, do nothing. You should keep the tree still a binary search tree after removal. 设找到的需要删除的节点为node - ...
elif node.right is None: # 只有左侧孩子节点 self.__reassign_nodes(node, node.left) else: # 左右孩子结点都有 tmp_node = self.get_max(node.left) # 找到左子树的最大结点 self.remove(tmp_node.value) # 删除节点 node.value = tmp_node.value # 不改变树结构,只更改当前节点的值 12.利用递归...
LintCode "Remove Node in Binary Search Tree" Not hard to find a solution, but there are several corner cases. classSolution {public:/** * @param root: The root of the binary search tree. * @param value: Remove the node with given value....
Deletion in a Binary Search Tree Example Here are the 4 steps that should be followed: 1) The node to be deleted is a leaf node: 2) The node to be deleted has only one child: 3) The node to be deleted has two children: Algorithm for Deletion in a Binary Search Tree Let's take...
// 向以node为根的二分搜索树中插入元素e,递归算法 private void add(Node node, E e){ if(e.equals(node.e)) return; else if(e.compareTo(node.e) < 0 && node.left == null){ node.left = new Node(e); size ++; return; }
parent = node; size ++; return newNode; } node = node.lesser; } else { if(node.greater == null) { node.greater = newNode; newNode.parent = node; size ++; return newNode; } node = node.greater; } } return newNode; } @Override public T remove(T value) { Node<T> node = ...
private void add(Node node, E e) { if (node.e.compareTo(e)) { // 不考虑重复元素 return; } else if (node.e.compareTo(e) > 0 && node.left == null) { node.left = new Node(e); size++; return; } else if (node.e.compareTo(e) < 0 && node.right == null) { ...
Binary tree (a) has 8 nodes, with node 1 as its root. Node 1's left child is node 2; node 1's right child is node 3. Notice that a node doesn't need to have both a left child and right child. In binary tree (a), node 4, for example, has only a right child. Furthermor...