[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.

 

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/**

 * 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 *DFS(vector<int> &preorder, vector<int> &inorder,int leftstart,int leftend,int rightstart,int rightend)

    {

        if(leftend-leftstart<0||rightend-rightstart<0) return NULL;

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

        int pos;//中序遍历中根结点位置

        for(int i=rightstart;i<=rightend;i++)

        {

            if(inorder[i]==preorder[leftstart]) 

            {

                pos=i;

                break;

            }

        }

        int len=pos-rightstart;

        root->left=DFS(preorder,inorder,leftstart+1,leftstart+len,rightstart,pos-1);

        root->right=DFS(preorder,inorder,leftstart+len+1,leftend,pos+1,rightend);

        return root;

    }

    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {

        return DFS(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);

    }

};