剑指Offer-7：重建二叉树

题目：

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。

Example:

例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

问题解析：

树存在先序遍历、中序遍历、后序遍历三种遍历方式。

• 先序遍历和中序遍历重建数组；
• 后序遍历和中序遍历重建数组。

链接：

剑指Offer（第2版）：P62

LeetCode：

思路标签：

算法：遍历、递归

数据结构：二叉搜索树

解答：

1. C++

• 先序遍历序列的第一个数字是根结点值；
• 中序遍历中根结点的左边序列是左子树，右边序列是右子树。
• 解题是特别还需要处理不符合条件的情况：序列为空的情况；两序列元素个数不同的情况；两序列中元素不相同的情况。

先序遍历和中序遍历：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> vin) {        if (pre.size() == 0 || vin.size() == 0)            return nullptr;        int *preStart = &pre[0];        int *preEnd = &pre[pre.size() - 1];        int *vinStart = &vin[0];        int *vinEnd = &vin[vin.size() - 1];        return ConstructCore(preStart, preEnd, vinStart, vinEnd);    }    TreeNode* ConstructCore(int* preStart, int* preEnd, int* vinStart, int* vinEnd)    {        int rootValue = preStart[0];        TreeNode* root = new TreeNode(preStart[0]);        root->val = rootValue;        root->left = nullptr;        root->right = nullptr;        if (preStart == preEnd)        {            if (vinStart == vinEnd && *preStart == *vinStart)                return root;            //else            //throw std::exception("Invalid input");        }        int *rootVin = vinStart;        while (rootVin < vinEnd && *rootVin != rootValue)            ++rootVin;        //if(rootVin == vinEnd && *rootVin != rootValue)        //throw std::exception("Invalid input");        int leftLength = rootVin - vinStart;        int* preLeftEnd = preStart + leftLength;        if (leftLength > 0)        {            //构建左子树            root->left = ConstructCore(preStart + 1, preLeftEnd, vinStart, rootVin - 1);        }        if (leftLength < preEnd - preStart)        {            //构建右子树            root->right = ConstructCore(preLeftEnd + 1, preEnd, rootVin + 1, vinEnd);        }        return root;    }    void preorder(TreeNode* root) {        if (root) {            cout << root->val;            preorder(root->left);            preorder(root->right);        }        cout << endl;    }};

中序遍历和后序遍历：

class Solution {public:    TreeNode* reConstructBinaryTree(vector<int> post, vector<int> in){        int postLength = post.size();        int inLength = in.size();        if (postLength == 0 || inLength == 0) return nullptr;        int* postStart = &post[0];        int* postEnd = &post[postLength - 1];        int* inStart = &in[0];        int* inEnd = &in[inLength - 1];        return ConstructCore(postStart, postEnd, inStart, inEnd);    }    TreeNode* ConstructCore(int* postStart, int* postEnd, int* inStart, int* inEnd)    {        int rootValue = *postEnd;        TreeNode* root = new TreeNode(rootValue);        if (postStart == postEnd) {            if (inStart == inEnd && *inStart == *postStart)                return root;            else                throw std::exception("Invalid value!");        }        int* inRoot = inStart;        while (inRoot < inEnd && *inRoot != rootValue)            ++inRoot;        if (inRoot == inEnd && *inRoot != rootValue)            throw std::exception("Invalid value!");        int rightLength = inEnd - inRoot;        int* rightPostStart = postEnd - rightLength;        if (rightLength > 0) {            root->right = ConstructCore(rightPostStart, postEnd - 1, inRoot + 1, inEnd);        }        if (rightLength < postEnd - postStart) {            root->left = ConstructCore(postStart, rightPostStart - 1, inStart, inRoot - 1);        }        return root;    }};

2. Java

先序遍历和中序遍历：

http://blog.csdn.net/koala_tree/article/details/78548628

中序遍历和后序遍历：

http://blog.csdn.net/koala_tree/article/details/78548058