LeetCode-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* Build(vector<int>& inorder,vector<int>& preorder,int s1,int e1,int s2,int e2){
if(s1>e1||s2>e2)return NULL;
TreeNode* ret=new TreeNode(preorder[s2]);
int i=s1;
while(i<=e1){
if(inorder[i]==preorder[s2])break;
i++;
}
ret->left=Build(inorder,preorder,s1,i-1,s2+1,s2+i-s1);
ret->right=Build(inorder,preorder,i+1,e1,s2+i-s1+1,e2);
}
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
// Note: The Solution object is instantiated only once and is reused by each test case.
return Build(inorder,preorder,0,inorder.size()-1,0,inorder.size()-1);
}
};

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