construct-binary-tree-from-preorder-and-inorder-traversal——前序和中序求二叉树

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()!=inorder.size()) return NULL;
        return build(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);
    }
    
    TreeNode *build(vector<int> &preorder, vector<int> &inorder,int s1,int e1,int s2,int e2){
        if(s1>e1||s2>e2) return NULL;
        if(s1==e1) return new TreeNode(preorder[s1]);
        TreeNode *res=new TreeNode(preorder[s1]);
        int tmp=preorder[s1];
        int index=0;
        while((s2+index)<=e2){
            if(inorder[s2+index]==tmp)
                break;
            index++;
        }
        res->left=build(preorder,inorder,s1+1,s1+index,s2,s2+index-1);
        res->right=build(preorder,inorder,s1+index+1,e1,s2+index+1,e2);
        return res;
    }
};