问题重述: 问题求解: 我们换一组比较有代表性的样例, 对于上图的树来说, index: 0 1 2 3 4 5 6 先序遍历为: 中序遍历为: 5同理,右子树也是如此.这样不难看出本题应该用递归方法解决. /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } *…
Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 可与Construct Binary Tree from Inorder and Postorder Trav…
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…
Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 给出前序遍历和中序遍历,然后求这棵树. 很有规律.递归就可以实现. /** * Definition for a binary tree node. * public class TreeNode { * int val; *…
Given preorder and inorder traversal of a tree, construct the binary tree. ============== 基本功: 利用前序和中序构建二叉树 , === code /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : va…
Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 通过树的中序序列和前序序列来构建一棵树. 例如:一棵树的前序序列是1-2-3,中序序列有5种可能,分别是1-2-3.2-1-3.1-3-2.2-3-1.3-2-1.不同的中序序列对应着不同的树的结构,这几棵树分别是[1,nul…
Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. Subscribe to see which companies asked this question     递归就可以了.   #include<algorithm> using namespace std; /**…
Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 题目标签:Array, Tree 题目给了我们preOrder 和 inOrder 两个遍历array,让我们建立二叉树.先来举一个例子,让我们看一下preOrder 和 inOrder的特性. / \   /      \…
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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…