之后在[inorder+index+1, end] [0, inorder+index-1] 分别构造右树和左树 intsearchNode(intinorder[],intinorderSize,intkey){inti;for(i=0;i<inorderSize;i++){if(key==inorder[i]){returni;}}return-1;}structTreeNode*buildTree(int*preorder,intpreorderSize,int*inorder,intinorderSize){...
right of the last node := i insert i into the stack return root Let us see the following implementation to get better understanding − Example classSolution(object):defbstFromPreorder(self,preorder):""" :type preorder: List[int] :rtype: TreeNode """root=TreeNode(preorder[0])stack=[...
TreeNode*root=NULL;intpreBegin=0,preEnd=preorder.size()-1;intinBegin=0,inEnd=inorder.size()-1; buildBST(root,preorder,preBegin,preEnd,inorder,inBegin,inEnd);returnroot; } };
node->right =bstFromPreorder(right);returnnode; } }; Github 同步地址: https://github.com/grandyang/leetcode/issues/1008 类似题目: Construct Binary Tree from Preorder and Postorder Traversal Construct Binary Tree from Inorder and Postorder Traversal Construct Binary Tree from Preorder and Inorder...
How to create BST by using Inorder and Postorder traversal? +1 vote Given inorder traversal of a binary tree. Print all the possible binary trees from it? +1 vote How to program Inorder traversal of Binary tree without stack and recursion? +1 vote Given preorder traversal of a BST...
**/fun bstFromPreorder(preorder: IntArray): TreeNode?{if(preorder ==null||preorder.isEmpty()) {returnnull} val size=preorder.size val root= TreeNode(preorder[0])for(i in 1until size) { help(root, preorder[i]) }returnroot ...
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: ...