left, right. postorder: left, right, root. 这时候你依然发现可以从左往右,preorder最左边就是root,所以用它来建立treenode,用postorder做参考,递归的时候就只传当前node就可以了,因为preorder是不断在建立left subtree,所以当preorder[i] == postorder[j] 的时候,你找到了最左边的treenode,然后返回,这时候返...
=postorder.length)returnnull;Map<Integer,Integer>map=newHashMap<Integer,Integer>();for(inti=0;i<inorder.length;i++){map.put(inorder[i],i);}intlen=inorder.length;returnhelper(inorder,0,len-1,postorder,0,len-1,map);}publicTreeNodehelper(int...
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...
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...
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(...
1. Problem Descriptions:Given two integer arrays inorderandpostorderwhereinorderis the inorder traversal of a binary tree andpostorderis the postorder traversal of the same tree, construct and retu…
postorder = [9,15,7,20,3] 1. 2. Return the following binary tree: 3 / \ 9 20 / \ 15 7 1. 2. 3. 4. 5. 题解: 同105 classSolution{ public: TreeNode*buildTree(intpleft,intpright,intileft,intiright,vector<int>&postorder,vector<int>&inorder) { ...
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
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...
""" Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20 / \ 15 7 ...