2、postorder + inorder 1/**2* Definition for binary tree3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12typedef vector<int>::iterator Iter;1314TreeNode *build(...
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { return createTree(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1); } };Construct Binary Tree from Preorder and Inorder TraversalGiven preorder and inorder traversal of a tree, construct the binary tr...
链接:https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/ 难度:Medium 题目:106. Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicat...
LeetCode: 889. Construct Binary Tree from Preorder and Postorder Traversal 题目描述 Return any binary tree that matches the given preorder and postorder traversals. Values in the traversals pre and post are distinct positive integers. Example 1: Input: pre =...
Tree from Preorder and Postorder Traversal./*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/classSolution{public:TreeNode*constructFromPrePost(vector<int>&pre,...
You may assume that duplicates do not exist in the tree. 思路跟之间拿到inorder + preorder construct binary tree基本是一样的思路。根据postorder里最后面的是root的规律,我们维持一个postIndex从后面往前依次遍历postorder. 每次指向某个postorder里的元素,那我们就构建一个以该元素的值为val的tNode. 再去in...
Return the root of the constructed binary tree. 3.1.2 Indices as parameters: Start by defining a helper function build_tree_helper that takes 4 parameters inorder and postorder's start and end respectively . Identify the root of the binary tree using the last element of the postorder list. ...
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思路: easy 算法: 1. public TreeNode buildTree(int[] inorder, int[] postorder) { 2. if (inorder == null || inorder.length == 0 || postorde...
889. Construct Binary Tree from Preorder and Postorder Traversal,程序员大本营,技术文章内容聚合第一站。
type Index struct{index int}funcbuildTree(inorder[]int,postorder[]int)*TreeNode{iflen(inorder)==0{returnnil}//后续遍历的最后一个元素,是二叉树的root节点rootIndex:=&Index{len(postorder)-1}returnbuildTreeR(inorder,postorder,0,len(postorder)-1,rootIndex)}funcbuildTreeR(inorder[]int,postorder...