root.right= self.buildTree(preorder[i+1:], inorder[i+1:])returnroot 同样的方法做106. Construct Binary Tree from Inorder and Postorder Traversal,后序和中序,建二叉树,根节点是后序的最后一个元素。 classSolution(object):defbuildTree(self, inorder, postorder):""":type inorder: List[int] ...
原题链接在这里:https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/ 题目: Return the root node of a binary search tree that matches the givenpreordertraversal. (Recall that a binary search tree is a binary tree where for everynode, any descendant ofnode.lefthas...
TreeNode* buildTree(vector<int>& preorder, int preBeg, vector<int>& inorder, int inBeg, int size) { if(size <= 0) return nullptr; TreeNode* curNode = new TreeNode(preorder[preBeg]); int inorderLeftTreeEnd = inBeg; while(inorder[inorderLeftTreeEnd] != preorder[preBeg]) ++in...
private TreeNode helper(int[] preorder, int preL, int preR, int[] inorder, int inL, int inR, HashMap<Integer, Integer> map) { if(preL>preR || inL>inR) return null; TreeNode root = new TreeNode(preorder[preL]); int index = map.get(root.val); root.left = helper(preorder, ...
You may assume that duplicates do not exist in the tree. 翻译:给定树的前序和中序遍历,构造二叉树。 注意: 树中不存在重复项。 思路:首先,你应该知道 前序遍历:根节点,左子树,右子树; 中序遍历:左子树,根节点,右子树; 所以,我们可以从preorder中找到整棵树的根节点,即为preorder[0],由于preorder和...
* TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{public TreeNodebuildTree(int[]preorder,int[]inorder){if(preorder==null||preorder.length==0)returnnull;if(inorder==null||inorder.length==0)returnnull;if(preorder.length!=inorder.length...
105. 从前序与中序遍历序列构造二叉树 - 给定两个整数数组 preorder 和 inorder ,其中 preorder 是二叉树的先序遍历, inorder 是同一棵树的中序遍历,请构造二叉树并返回其根节点。 示例 1: [https://assets.leetcode.com/uploads/2021/02/19/tree.jpg] 输入: preo
Construct Binary Tree from Preorder and Inorder Traversal 题目描述(中等难度) 根据二叉树的先序遍历和中序遍历还原二叉树。 解法一 递归 先序遍历的顺序是根节点,左子树,右子树。中序遍历的顺序是左子树,根节点,右子树。 所以我们只需要根据先序遍历得到根节点,然后在中序遍历中找到根节点的位置,它的左边就...
* 题目: 105.Construct Binary Tree from Preorder and Inorder Traversal * 网址:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ * 结果:AC * 来源:LeetCode * 博客: ---*/#include<iostream>#include<vector>#include<algorithm>usingnamespacestd;structTree...
题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 根据前序遍历和中序遍历结果构造二叉树。 思路分析: 分析二叉树前序遍历和中序遍历的结果我们发现: 二叉树前序遍历的第一个节点是根节点。 在中序遍历...