105. 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.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

方法1:

basketking：https://www.youtube.com/watch?v=S1wNG5hx-30

思路：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution1 {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int preStart = 0;
        int inStart = 0;
        int inEnd = preorder.size() - 1;
        if (preorder.size() == 0) return NULL;
        
        return helper(preorder, inorder, preStart, inStart, inEnd);
    }
    
    TreeNode* helper(vector<int> & preorder, vector<int> & inorder, int preStart, int inStart, int inEnd){
        // ending case is inStart > inEnd
        // happen when the child node is null
        // current will create the leaf node;
        if (preStart >= preorder.size()|| inStart > inEnd) return NULL;
       
        
        TreeNode* current = new TreeNode(preorder[preStart]);
       
        int k;
         
        for (int i = inStart; i <= inEnd; i++){
            if (preorder[preStart] == inorder[i]){
                k = i;
                break;  
            }
        }
        // i is the position of preorder[preStart] in inorder
        
        current->left = helper(preorder, inorder, preStart + 1, inStart, k - 1);
        current->right = helper(preorder, inorder, preStart + (k - inStart) + 1, k + 1, inEnd);
        return current;
    }
};