1TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {2//Start typing your C/C++ solution below3//DO NOT write int main() function4TreeNode *root =newTreeNode(0);5if(inorder.size() ==0){6returnNULL;7}8vector<int>leftInorder, leftPostorder, rightInorder, rightPostor...
标志31if(startPostorder==endPostorder){32returnnewTreeNode(postorder[endPostorder]);33}else{34//记录根结点的在中序遍历中的位置35introotIn=startInorder;36for(inti=startInorder;i<=endInorder;i++){37if(inorder[i]==postorder[endPostorder]){38rootIn=i;39break;40}41}42//创建根结点43TreeNod...
TreeNode *buildTree(vector<int> &inorder,vector<int> &postorder){ returnhelper(inorder,0, inorder.size() -1, postorder,0, postorder.size() -1); } private: TreeNode *helper(vector<int> &inorder,intinStart,intinEnd,vector<int> &postorder,intpostStart,intpostEnd){ if(inStart > inEnd...
LeetCode—106. Construct Binary Tree from Inorder and Postorder Traversal,程序员大本营,技术文章内容聚合第一站。
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: 这题的思路与 105 Construct Binary Tree from Preorder and Inorder Traversal 基本相同。
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。
Given inorder and postorder traversal of a tree, construct the binary tree. 根据树的中序(左根右)和后序遍历(左右根)构造一棵二叉树· 后序遍历最后一个元素是根节点,通过根节点将中序遍历数组分为两个数组,分别是左子树和右子树节点的集合,在进行递归调用 ...
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. ...
Can you solve this real interview question? Construct Binary Tree from Inorder and Postorder Traversal - Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of th
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 翻译:给定树的中序遍历和后序遍历,构造二叉树。 注意:树中不存在重复项。 思路:本题与105. Construct Binary Tree from Preorder and Inorder Traversal类似。