//226. Invert Binary Tree 反转二叉树//思路:递归互换左右子节点//Example:///Input: Output:///4 4/// \ / \//2 7 7 2/// \ / \ / \ / \//1 3 6 9 9 6 3 1func invertTree(root *TreeNode) *TreeNode {ifroot ==nil {returnroot } root.Left, root.Right=invertTree(root.Righ...
varans[]stringfuncbinaryTreePaths(root*TreeNode)[]string{ans=make([]string,0)dfs(root,"")returnans}funcdfs(root*TreeNode,pathstring){ifroot==nil{return}path=fmt.Sprintf("%s%d",path,root.Val)ifroot.Left==nil&&root.Right==nil{ans=append(ans,path)return}path+="->"dfs(root.Left,path)...
var rootNode *BinaryTree = &BinaryTree{ Value: 1, LeftNode: &BinaryTree{ Value: 2, LeftNode: &BinaryTree{ Value: 4, RightNode: &BinaryTree{ Value: 6, LeftNode: &BinaryTree{Value: 7}, RightNode: &BinaryTree{Value: 8}, }, }, }, RightNode: &BinaryTree{ Value: 3, RightNode: ...
type BinaryTree struct { Value int Left *BinaryTree Right *BinaryTree } func InitBinaryTree(root *BinaryTree) *BinaryTree { l := BinaryTree{} r := BinaryTree{} root.Left = l.NewBinaryTreeNode(2) root.Right = r.NewBinaryTreeNode(3) l2:=BinaryTree{} ll2:=BinaryTree{} root.Left....
// 定义二叉树typeBinaryTreestruct{DataintLeft*BinaryTreeRight*BinaryTree}// 递归反转funcInvertBinaryTree(root*BinaryTree)*BinaryTree{// 递归退出条件ifroot==nil{returnnil}// 反转左右结点的值root.Left,root.Right=root.Right,root.Left// 递归左子树InvertBinaryTree(root.Left)// 递归右子树InvertBinary...
156.二叉树的上下翻转 Binary Tree Upside Down 给定一个二叉树,满足所有右节点要么是叶子节点,要么没有左兄弟节点,将它上下颠倒并转化为一个树,原来的右节点变成了左叶子节点。返回新的根节点。 例如,给定 binary tree [1,2,3,4,5] 为: 1/ \2 3/ \4 5 ...
完全二叉树(Complete Binary Tree)若设二叉树的深度为h，除第 h 层外，其它各层 (1～h-1) 的结点数都达到最大个数，第 h 层所有的结点都连续集中在最左边，这就是完全二叉树。完全二叉树是由满二叉树而引出来的。对于深度为K的，有n个结点的二叉树，当且仅当其每一个结点都与深度为K的满二叉树中编号...
e:=Node{nil,nil,"5"}d:=Node{nil,nil,"4"}c:=Node{left:nil,right:&e,value:"3"}b:=Node{nil,nil,"6"}a:=Node{left:&c,right:&d,value:"2"}head:=Node{left:&a,right:&b,value:"1"}BinaryTreePath(&head,path)fmt.Println(res)}funcBinaryTreePath(root*Node,path[]string){path=...
v2=<-ch2ifv1!=v2{returnfalse}}returntrue}funcmain(){fmt.Println(Same(tree.New(1),tree.New(1)))fmt.Println(Same(tree.New(1),tree.New(2)))}EOF #NOTE:提示,缺少go.mod 文件[root@sltvb7v2wy3 jia_concurrency]# gor exercise_equivalent_binary_trees.goexercise_equivalent_binary_trees.go:...
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func inorderTraversal(root *TreeNode) []int { var m []int var s Stack cur:=root for !s.empty() || cur!=nil{ for cur!=nil{ s.push(cur) cur=cur....