publicclassTreeNode{intval;TreeNodeleft;TreeNoderight;TreeNode(intval=val; 基本概念 "二叉树"(Binary Tree)这个名称的由来是因为二叉树的每个节点最多有两个子节点,一个左子节点和一个右子节点。其中,“二叉”指的是两个,因此“二叉树”表示每个节点最多可以分支成两个子节点。基本定义: 每个节点
一般来说,节点多指计算机网络中的物理节点,而算法中的Node通常译为结点。 结点的度(degrees):结点子树的个数。 结点的层(levels):如果根结点的层定义为1(有时候定义为0),根的子结点为第2层结点,以此类推。 树的深度:树中最大的结点层。 满二叉树(Full Binary Tree):除叶子结点的度为0外,其余所有结点的度...
一棵树6BinaryTree binTree =newBinaryTree();7//创建一个根节点8TreeNode root =newTreeNode(1);9//给树对象赋值10binTree.setRoot(root);11//创建左右两个子节点12TreeNode rootL =newTreeNode(2);13TreeNode rootR =newTreeNode(3);14//将创建的子节点的引用传递给根节点15root...
public void setLeft(BinaryTreeNode<T> left) { this.left = left; } public void insertLeft(T data) { setLeft(new BinaryTreeNode<T>(data)); } public void setRight(BinaryTreeNode<T> right) { this.right = right; } public void insertRight(T data) { setRight(new BinaryTreeNode<T>(da...
1. Maximum Depth of Binary Tree 解法1: class Solution: def maxDepth(self, root: TreeNode) -> int: self.answer = 0 self.helper(root,0) return self.answer def helper(self,node,depth): if node is None: self.answer=max(self.answer,depth) else: self.helper(node.left,depth+1) self....
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool checkSubTree(TreeNode* t1, TreeNode* t2) { if(!t2) return true...
* var ti = TreeNode(5) * var v = ti.`val` * Definition for a binary tree node. * class TreeNode(var `val`: Int) { * var left: TreeNode? = null * var right: TreeNode? = null * } *//** * Queue Solution */funlevelOrder(root:TreeNode?):List<List<Int>>{val ans=mutabl...
Thedegree of a treeis the maximum degree of the nodes in the tree. A node with degree zero is aleaforterminalnode. A node that has subtrees is theparentof the roots of the subtrees, and the roots of the subtrees are thechildrenof the node. ...
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...
insertTreeNode=(root,v)=>{letqueue=[root]while(true){varnode=queue.pop()if(!node.value){node.value=vbreak}if(!node.left){node.left={value:v}break}else{queue.unshift(node.left)}if(!node.right){node.right={value:v}break}else{queue.unshift(node.right)}}console.log('tree',root)} ...