TreeNode* create(vector<int>& inorder, vector<int>& postorder,intis,intie,intps,intpe) {if(ps > pe ||is> ie)returnnullptr; TreeNode* node =newTreeNode(postorder[pe]);intpos;for(inti =is; i <= ie; ++i) {if(inorder[i] == node->val) { pos=i;break; } } node->left = ...
left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {13if(inorder.size()!=postorder.size()||inorder.size()<1)14returnNULL;15returnbuild(inorder,
与Construct Binary Tree from Inorder and Preorder Traversal问题非常类似,唯一区别在于这一次确定root的位置由post traversal来确定,为最后一个元素。 AI检测代码解析 1/**2* Definition for binary tree3* public class TreeNode {4* int val;5* TreeNode left;6* TreeNode right;7* TreeNode(int x) { ...
for(int i = 0; i < inorder.length; i ++) map.put(inorder[i], i); return helper(inorder, 0, inorder.length - 1, postorder, postorder.length - 1, map); } TreeNode helper(int[] inorder, int startIn, int endIn, int[] postorder, int endPost, HashMap<Integer, Integer> map)...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution{public:TreeNode*buildTree(vector<int>&inorder,vector<int>&postorder){if(inorder.size()==0)returnNULL;intsize=postorder.size();TreeNode*root=newTreeNode(postorder[size-1]);postorder.erase(postorder.end(...
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. *//* MARK: - 题目翻译: 给一棵树的的中序遍历 和 后序遍历,构造该二叉树。 注: 您可以假设树中不存在重复项。
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
preorder(root); printf("\ninorder traversal of tree\n"); inorder(root); printf("\npostorder traversal of tree\n"); postorder(root); break; case2: insert(); printf("\npreorder traversal of tree\n"); preorder(root); printf("\nInorder traversal of tree\n"); ...
Return the root of the constructed binary tree. 3.1.2.1 Tips of finding the boundary of inorder and postorder: The inorder_index partitions the inorder list into two parts, with neither part including the index itself. Therefore, for the left subtree, the range is from the start index to...
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。