数据结构——根据中序遍历与先序遍历构建二叉树

原文地址：Construct Tree from given Inorder and Preorder traversals

我们考虑下下面的遍历：

中序遍历：D B E A F C
先序遍历：A B D E C F

在一个先序序列中，最左端的元素就是树根。所以我们知道A是已知序列的根。通过查询A的中序序列，我们可以得到A左边左子树的所有元素和右边右子树的所有元素。所以我们现在知道了以下结构。

                 A               /   \             /       \           D B E     F C

我们递归上述步骤，然后得到下面的树：

         A       /   \     /       \    B         C   / \        / /     \    /D       E  F

算法： buildTree()

1）从先序序列中选一个元素。增加一个先序索引变量（下面代码中的preIndex）来选择下一次递归调用的下一个元素。
2）用已选择的元素的值创建一个新的树节点tNode。
3）找到已选择元素在中序遍历中的索引。设这个索引为inIndex。
4）在inIndex之前调用buildTree，建立树作为tNode的左子树。
5）在inIndex之后调用buildTree，建立树作为tNode的右子树。
6）返回tNode。

// Java program to construct a tree using inorder and preorder traversal/* A binary tree node has data, pointer to left child   and a pointer to right child */class Node {    char data;    Node left, right;    Node(char item)     {        data = item;        left = right = null;    }}class BinaryTree {    Node root;    static int preIndex = 0;    /* Recursive function to construct binary of size len from       Inorder traversal in[] and Preorder traversal pre[].       Initial values of inStrt and inEnd should be 0 and len -1.         The function doesn't do any error checking for cases where        inorder and preorder do not form a tree */    Node buildTree(char in[], char pre[], int inStrt, int inEnd)     {        if (inStrt > inEnd)             return null;        /* Pick current node from Preorder traversal using preIndex           and increment preIndex */        Node tNode = new Node(pre[preIndex++]);        /* If this node has no children then return */        if (inStrt == inEnd)            return tNode;        /* Else find the index of this node in Inorder traversal */        int inIndex = search(in, inStrt, inEnd, tNode.data);        /* Using index in Inorder traversal, construct left and           right subtress */        tNode.left = buildTree(in, pre, inStrt, inIndex - 1);        tNode.right = buildTree(in, pre, inIndex + 1, inEnd);        return tNode;    }    /* UTILITY FUNCTIONS */    /* Function to find index of value in arr[start...end]     The function assumes that value is present in in[] */    int search(char arr[], int strt, int end, char value)     {        int i;        for (i = strt; i <= end; i++)         {            if (arr[i] == value)                return i;        }        return i;    }    /* This funtcion is here just to test buildTree() */    void printInorder(Node node)     {        if (node == null)            return;        /* first recur on left child */        printInorder(node.left);        /* then print the data of node */        System.out.print(node.data + " ");        /* now recur on right child */        printInorder(node.right);    }    // driver program to test above functions    public static void main(String args[])     {        BinaryTree tree = new BinaryTree();        char in[] = new char[]{'D', 'B', 'E', 'A', 'F', 'C'};        char pre[] = new char[]{'A', 'B', 'D', 'E', 'C', 'F'};        int len = in.length;        Node root = tree.buildTree(in, pre, 0, len - 1);        // building the tree by printing inorder traversal        System.out.println("Inorder traversal of constructed tree is : ");        tree.printInorder(root);    }}// This code has been contributed by Mayank Jaiswal

输出：

Inorder traversal of constructed tree is :D B E A F C

时间复杂度：O(n2)，当树是一个左退化树的时候，发生最坏的情况。例如先序遍历与中序遍历最坏的情况是 {A, B, C, D}与{D, C, B, A}。