剑指 Offer 07. 重建二叉树

思路：模拟人工建树过程，先从前序遍历序列中找父节点，以父节点将中序遍历序列分为两半，可以找到左右子树长度，再对左右子树重复以上过程。

/**
 * 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:
    map<int, int> hash;

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int n = preorder.size();
        for(int i=0; i<n; i++) hash[inorder[i]] = i;
        return dfs(preorder, inorder, 0, n-1, 0, n-1);
    }

    TreeNode* dfs(vector<int>& preorder, vector<int>& inorder, int pl, int pr, int il ,int ir){
        if(pl > pr) return NULL;

        TreeNode* root = new TreeNode(preorder[pl]);

        int k = hash[root->val] - il;//区间长度

        TreeNode* left = dfs(preorder, inorder, pl+1, pl + k, il, il + k);

        TreeNode* right = dfs(preorder, inorder, pl + 1 + k, pr, il + k + 1, ir);

        root -> left = left;
        root -> right = right;
        
        return root;
    }
};