[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);
    }
};