private Node root; // 添加数据 public void add(int data) { // 递归调用 if (null == root) root = new Node(data, null, null); else addTree(root, data); } private void addTree(Node rootNode, int data) { // 添加到左边 if (rootNode.data > data) { if (rootNode.left == null...
}//向以node为根的二分搜索树中插入元素e,递归算法//返回插入新节点后二分搜索树的根privateNode add(Node node, E e) {if(node ==null) { size++;returnnewNode(e); }if(e.compareTo(node.e) < 0) { node.left=add(node.left, e); }elseif(e.compareTo(node.e) > 0) { node.right=add...
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; } else if(e.compareTo(node.e) > 0 && node.right == null){ node.right = new Node(e); size ++; retur...
importsysimportioclassSolution:def__init__(self):self.min_node=sys.maxsize+1self.res_path=[]self.tree=[]defsmallest_node(self,input_values:list)->list:self.tree=input_values# Find min and add -1 to all empty child nodesforiinrange(len(self.tree)):if0<self.tree[i]<self.min_node:...
publicList<Integer>inorderTraversal3(TreeNoderoot){List<Integer>ans=newArrayList<>();TreeNodecur=root;while(cur!=null){//情况 1if(cur.left==null){ans.add(cur.val);cur=cur.right;}else{//找左子树最右边的节点TreeNodepre=cur.left;while(pre.right!=null&&pre.right!=cur){pre=pre.right;}...
""" Definition of TreeNode: class TreeNode: def __init__(self, val): this.val = val this.left, this.right = None, None """ class Solution: """ @param root: The root of the binary search tree. @param node: insert this node into the binary search tree. @return: The root of...
(ans,0,root)returnans}funvisitLevel(ans:MutableList<MutableList<Int>>,level:Int,node:TreeNode?){if(null==node)return// level 从0 (root) 开始,此时 ans.size = 0; 每层的值存在 levelList 中. 这地方的代码非常巧妙.if(ans.size==level){val levelList=mutableListOf<Int>()ans.add(levelList...
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. Further...
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...
Summary: Nested | Field | Constr | Method Detail: Field | Constr | Method SEARCH: Module jdk.compiler Package com.sun.source.tree Interface BinaryTree All Superinterfaces: CaseLabelTreePREVIEW, ExpressionTree, Tree public interface BinaryTree extends ExpressionTree A tree node for a binary ...