[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.

 

方法一：最先想到的就是递归，注意low、high的计算，还有初始时NULL情况的处理。。

从inorder中寻找pre的第一个，然后左右递归

class Solution
{
    public:
        TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder)
        {
            if(preorder.size() == 0 || inorder.size() == 0)
                return NULL;
            return buildTree(preorder, 0, preorder.size()-1,
                              inorder, 0, inorder.size()-1);
        }     
              
        TreeNode *buildTree(vector<int> &preorder, int low1, int high1,
                            vector<int> &inorder, int low2, int high2)
        {     
            //cout << "==============" <<endl;
            //cout << "low1 = \t" << low1 <<endl;
            //cout << "high1= \t" << high1 <<endl;
            //cout << "low2 = \t" << low2 <<endl;
            //cout << "high2= \t" << high2 <<endl;
              
            TreeNode * p = new TreeNode(preorder[low1]);
            if(low1 == high1)
            {   
                return p;
            }   
            int index = 0;
            for(index = low2; index < high2; index++)
            {   
                if(inorder[index] == preorder[low1])
                    break;
            }
            //cout << "index= \t" << index<<endl;

            if(index != low2)
                p->left = buildTree(preorder, low1+1,(low1+1) + (index-1-low2), inorder, low2, index-1);
            if(index != high2)
                p->right = buildTree(preorder, high1 - (high2-index-1) ,high1, inorder, index+1, high2);

            return p;
        }
} ;

方法二：迭代