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.

class Solution {
public:
TreeNode * buildTree(vector<int>& preorder, int preBegin, int preEnd, vector<int>& inorder, int inBegin, int inEnd)
{
TreeNode*root =NULL;

if (preEnd >= preBegin)
{
root = new TreeNode(preorder[preBegin]);
int n = inEnd-inBegin+1;
int rootIndex = 0;
for (int i = inBegin; i <= inEnd; ++i)
{
if (inorder[i] == preorder[preBegin])
{
rootIndex = i;
break;
}
}

if (rootIndex!=inBegin)
{
root->left = buildTree(preorder, preBegin+1, preBegin+(rootIndex-inBegin), inorder, inBegin, rootIndex-1);
}
if (rootIndex!=inEnd)
{
root->right = buildTree(preorder, preBegin+1+(rootIndex-inBegin), preEnd, inorder, rootIndex+1, inEnd);
}
}
return root;
}

TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
return buildTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
}
};

注意：在利用递归算法解题时要注意节省内存。最初觉得通过传递下标参数计算麻烦，而试图每次重新构造vector数据进行参数传递，如下所示：

if (rootIndex!=0)
{
vector<int>p(preorder.Begin()+1, preorder.Begin()+rootIndex+1);
vector<int>q(inorder.Begin(), inorder.Begin()+rootIndex);
root->left = buildTree(p,q);
}

这么做有两个问题：

1. 在用迭代器初始化vector时，第二个迭代器要指向有效数据的下一个数；

2. 容易造成内存溢出的问题。（在leetcode中运行时就出现这个问题）。