root.right= self.buildTree(preorder[i+1:], inorder[i+1:])returnroot 同样的方法做106. Construct Binary Tree from Inorder and Postorder Traversal,后序和中序,建二叉树,根节点是后序的最后一个元素。 classSolution(object):defbuildTree(self, inorder, postorder):""":type inorder: List[int] ...
Given preorder and inorder traversal of a tree, construct the binary tree.解决思路首先确定根节点,然后确定左右子树的节点数目。依次递归即可。假设输入的序列均合法。程序1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 ...
TreeNode* buildTree(vector<int>& preorder, int preBeg, vector<int>& inorder, int inBeg, int size) { if(size <= 0) return nullptr; TreeNode* curNode = new TreeNode(preorder[preBeg]); int inorderLeftTreeEnd = inBeg; while(inorder[inorderLeftTreeEnd] != preorder[preBeg]) ++in...
return helper(preorder,0,preorder.length-1,inorder,0,inorder.length-1, map); } private TreeNode helper(int[] preorder, int preL, int preR, int[] inorder, int inL, int inR, HashMap<Integer, Integer> map) { if(preL>preR || inL>inR) return null; TreeNode root = new TreeNode...
4、Given preorder andinordertraversal of a tree, construct the binary tree. 给定一个二叉树的前序和中序遍历,重建这棵二叉树。 5、So,inorderto the health of everyone, please stop smoking. 所以,为了大家的健康,请戒烟。 6、He is cautiousinorderto get promoted. ...
The classic example was to construct a binary tree given its inorder and preorder traversal =-=[38, 39, 24, 83, 78]-=-. Inversion of functional programs has received even less attention. Most published results (e.g. [55, 40]) are based on a “compositional” approach, which is ...
* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicTreeNodebuildTree(int[]preorder,int[]inorder){if(preorder==null||inorder==null||preorder.length!=inor...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 翻译:给定树的前序和中序遍历,构造二叉树。 注意: 树中不存在重复项。 思路:首先,你应该知道 前序遍历:根节点,左子树,右子树; ...
Construct Binary Tree from Preorder and Inorder Traversal 题目描述(中等难度) 根据二叉树的先序遍历和中序遍历还原二叉树。 解法一 递归 先序遍历的顺序是根节点,左子树,右子树。中序遍历的顺序是左子树,根节点,右子树。 所以我们只需要根据先序遍历得到根节点,然后在中序遍历中找到根节点的位置,它的左边就...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 做了第106题再来做这个就简单了,写递归的话思路一模一样: /** * Definition for a binary tree node. ...