Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思路:由前序遍历数组和中序遍历数组构建二叉树,此题与Construct Binary Tree from Inorder and Postorder Traversal很相似。 1.从前序遍历找树的根结点,即第一...
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...
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...
intsearchNode(intinorder[],intinorderSize,intkey){inti;for(i=0;i<inorderSize;i++){if(key==inorder[i]){returni;}}return-1;}structTreeNode*buildTree(int*preorder,intpreorderSize,int*inorder,intinorderSize){if(preorder==NULL||inorder==NULL||preorderSize==0||inorderSize==0)return...
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 self.topLeft=topLeft self.topRight=topRight self.bottomLeft=bottomLeft ...
- 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...
这个题目的思路就是利用preorder 和inorder的两种特性, 我们可以发现, preorder[0] 总是root, 然后inorder 里面根据之前的root, 可以将inorder分为left tree和right tree, 然后return root, recursive call即可. 1. Constraints 1) pre or in can be empty ...
* 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) { ...