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] ...
Given preorder and inorder traversal of a tree, construct the binary tree.解决思路首先确定根节点,然后确定左右子树的节点数目。依次递归即可。假设输入的序列均合法。程序1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 ...
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...
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { //step1:construct hash-tab if(preorder.size()==0) return NULL; unordered_map<int,int> mapIndex; for(int i=0;i<inorder.size();i++) mapIndex[inorder[i]] = i; return helpTree(preorder,0,inorder.size(),0,m...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: 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; ...
TreeNode root=newTreeNode(preorder[preorderBegin]);int rootIndex=0;for(int i=inorderBegin;i<=inorderEnd;i++){if(inorder[i]==preorder[preorderBegin]){rootIndex=i;break;}}int leftTreeSize=rootIndex-inorderBegin;root.left=buildTree(preorderBegin+1,preorderBegin+leftTreeSize,inorderBegin...
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...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume thatduplicates do not existin the tree. 思路 主要是根据前序遍历和中序遍历的特点解决这个题目。 1、确定树的根节点。树根是当前树中所有元素在前序遍历中最先出现的元素。
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def buildTree(self, preorder, inorder): """ :type preorder: List[int] :type inorder: List[int] :...
105. Construct Binary Tree from Preorder and Inorder Traversal——tree,程序员大本营,技术文章内容聚合第一站。