Construct Binary Tree from Preorder and Inorder Traversal 前序和中序重建二叉树

Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.
Note:

You may assume that duplicates do not exist in the tree.

/**
 * 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 *buildTree(vector<int> &preorder, vector<int> &inorder) {
        
        if(preorder.size()==0|| inorder.size()==0)
            return NULL;
        return solve(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
    }
    
    TreeNode *solve(vector<int> &preorder,int s1,int e1,vector<int> &inorder,int s2,int e2)
    {
        int rootVal=preorder[s1];
        int rootIndex=-1;
        for(int i=s2;i<=e2;i++)
            if(inorder[i]==rootVal)
            {
                rootIndex=i;
                break;
            }
        int left=rootIndex-s2;
        int right=e2-rootIndex;
        TreeNode *rootNode=new TreeNode(rootVal);
        if(left>0)
            rootNode->left=solve(preorder,s1+1,s1+left,inorder,s2,rootIndex-1);
        if(right>0)
            rootNode->right=solve(preorder,s1+left+1,e1,inorder,rootIndex+1,e2);
    
        return rootNode;
    }
};