construct-binary-tree-from-preorder-and-inorder

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||preorder.size()!=inorder.size())
return NULL;
return recurbuild(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);
}
TreeNode *recurbuild(vector<int> &preorder,vector<int> &inorder,int prebegin,
int preend,int inbegin,int inend){
if(prebegin>preend)
return NULL;
TreeNode* root=new TreeNode(preorder[prebegin]);
auto pos=find(inorder.begin(),inorder.end(),preorder[prebegin]);
int len=pos-inorder.begin();
root->left=recurbuild(preorder,inorder,prebegin+1,prebegin+len-inbegin,inbegin,len-1);
root->right=recurbuild(preorder,inorder,prebegin+len-inbegin+1,preend,len+1,inend);
return root;
}
};