leetcode105. 根据前序遍历和中序遍历构造二叉树

题目链接：https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

根据前序遍历和中序遍历构造二叉树

思路：

1. 根据前序遍历确定根结点，哈希根节点的值，得到中序遍历下标。
2. 根据中序遍历下标，得到左子树大小。
3. 从而确定前序遍历和中序遍历中左子树，右子树开始的位置。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    unordered_map<int, int> index;
    TreeNode* build(vector<int>& preorder, vector<int>& inorder, int preorder_left, int preorder_right,int inorder_left,int inorder_right)
    {
        if(preorder_left>preorder_right)
        {
            return nullptr;
        }
        int root_val=preorder[preorder_left];
        int root_inorder=index[root_val];
        TreeNode *root= new TreeNode(root_val);
        int size_left = root_inorder - inorder_left;//左子树大小
        root->left=build(preorder, inorder, preorder_left+1, preorder_left+size_left, inorder_left, inorder_left+size_left-1);
        root->right=build(preorder,inorder,preorder_left+size_left+1,preorder_right,root_inorder+1,inorder_right);
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int n = inorder.size();
        for(int i=0;i<inorder.size();i++)
        {
            index[inorder[i]]=i;
        }
        return build(preorder, inorder, 0, n-1, 0,n-1);
    }
};