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...
self.build(node.right,right)defbstFromPreorder(self, preorder):""":type preorder: List[int] :rtype: TreeNode"""root=TreeNode(preorder.pop(0)) self.build(root,preorder)returnroot
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...
1 <= preorder.length <= 100 The values ofpreorderare distinct. 题解: The first element should be root. As BST, root left subtree should be smaller than root value, right subtree should be bigger than root value. Could use root value as pivot and find out array corresponding to left ...
1classSolution {2func bstFromPreorder(_ preorder: [Int]) -> TreeNode?{3returnbuildTree(0, preorder.count-1, preorder)4}56func buildTree(_ start: Int, _ end: Int, _ preorder: [Int]) -> TreeNode?{7ifstart > end {returnnil }8ifstart == end {returnTreeNode(preorder[start])...
Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example,
* 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; ...