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.youkuaiyun.com/koala_tree/article/details/78548628

中序遍历和后序遍历：

http://blog.youkuaiyun.com/koala_tree/article/details/78548058