查询index部分可以使用Map做查询,建议和下一道题 leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 中后序构造BST和leetcode 449. Serialize and Deserialize BST 二叉搜索树BST的序列化和反序列化一起学习。 建议和leetcode 87. Scramble String 字符串拼凑 && DFS深度优先搜索 和 leetco...
6* public var right: TreeNode?7* public init(_ val: Int) {8* self.val = val9* self.left = nil10* self.right = nil11* }12* }13*/14classSolution {15func bstFromPreorder(_ preorder: [Int]) -> TreeNode?
node->right =bstFromPreorder(right);returnnode; } }; Github 同步地址: https://github.com/grandyang/leetcode/issues/1008 类似题目: Construct Binary Tree from Preorder and Postorder Traversal Construct Binary Tree from Inorder and Postorder Traversal Construct Binary Tree from Preorder and Inorder...
Leetcode 427. Construct Quad Tree 2. Solution **解析:**采用递归的方法构建Quad Tree,逐步分解即可。 Version 1 代码语言:javascript 代码运行次数:0 """ # Definitionfora QuadTree node.classNode:def__init__(self,val,isLeaf,topLeft,topRight,bottomLeft,bottomRight):self.val=val self.isLeaf=isLeaf ...
- 测试代码:funcConstructBinaryTreeFromInorderandPostorderTraversal(){letinorder=[4,2,5,1,6,3,7]letpostorder=[4,5,2,6,7,3,1]letroot=Solution().buildTree(inorder,postorder)print(root??"二叉树为空",CreateBinaryTree().isValidBST(root:root))//额外判断一下一颗二叉树是否为二叉查找树, 并...
right=preorder[len(left):]iflen(left) >= 1: node.left=TreeNode(left.pop(0)) self.build(node.left,left)iflen(right) >= 1: node.right=TreeNode(right.pop(0)) self.build(node.right,right)defbstFromPreorder(self, preorder):""":type preorder: List[int] ...
1/**2* Definition for a binary tree node.3* public class TreeNode {4* int val;5* TreeNode left;6* TreeNode right;7* TreeNode(int x) { val = x; }8* }9*/10classSolution {11publicTreeNode bstFromPreorder(int[] preorder) {12if(preorder ==null|| preorder.length == 0){13...
1) pre or in can be empty 2. ideas recursively DFS, 分别得到root, root.left & root.right, 返回root 3. Code classSolution:defconstructBT(self, preorder, inorder):ifnotpreorderornotinorder:return#note 别忘了not before inorderroot, index =TreeNode(preorder[0]), inorder.index(preorder...
* preorder:root->left->right,so the first element of preorder array is the root of tree **/fun bstFromPreorder(preorder: IntArray): TreeNode?{if(preorder ==null||preorder.isEmpty()) {returnnull} val size=preorder.size val root= TreeNode(preorder[0])for(i in 1until size) { ...
classSolution {int[] preorder;intidx = 0; Map<Integer, Integer> map_inorder =newHashMap<>();publicTreeNode bstFromPreorder(int[] preorder) {this.preorder =preorder;int[] inorder =Arrays.copyOf(preorder, preorder.length); Arrays.sort(inorder);intn =inorder.length;for(inti = 0; ...