6* public var right: TreeNode?7* public init(_ val: Int) {8* self.val = val9* self.left = nil10* self.right = nil11* }12* }13*/14classSolution {15vari =01617func 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...
/// M_106_ConstructBinaryTreeFromInorderandPostorderTraversal.swift// AlgorithmLeetCode/// Created by okerivy on 2017/3/18.// Copyright © 2017年 okerivy. All rights reserved.// https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversalimportFoundation// MARK: ...
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] :rtype: TreeNode"""root=...
链接:https://leetcode.cn/problems/construct-binary-search-tree-from-preorder-traversal 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 这一类比如 input 给的是树的某种遍历的结果,然后请你恢复整棵树的题目,都偏分治 + 递归的思路,这道题也不例外。题目给的是一棵 BST 的前序遍历...
(nr==-1&& preorder[i]>root->val)20nr =i;21}22if(nr==-1)23{24root->right =NULL;25root->left = solve(preorder,l+1,r);26returnroot;27}28root -> left = solve(preorder,l+1,nr-1);29root -> right =solve(preorder,nr,r);30returnroot;31}32TreeNode* bstFromPreorder(vector...
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...
classSolution {int[] preorder;intidx = 0;intn;publicTreeNode bstFromPreorder(int[] preorder) {this.preorder =preorder; n=preorder.length;returnhelper(Integer.MIN_VALUE, Integer.MAX_VALUE); }publicTreeNode helper(intlower,intupper){if(idx == n)returnnull;intval =preorder[idx];if(val...
* 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) { ...