72_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.

1:注意特殊情况。2:递归的情况；3:递归结束情况；4:首先获得根节点，之后把两个数组分别分成两部分，递归分别得出左子树和右子树。

TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder)
{
if(preorder.size() == 0 || inorder.size() == 0 || preorder.size() != inorder.size())
{
return NULL;
}

int size = (int)preorder.size();

return buildTreeCore(preorder, 0, inorder, 0, size);
}

TreeNode* buildTreeCore(vector<int> &preorder, int preStart,  vector<int> &inorder, int preIn, int length)
{
if(length == 1)
{
if(preorder[preStart] != inorder[preIn])
{
return NULL;
}
}

int rootValue = preorder[preStart];
TreeNode *root = new TreeNode(rootValue);

int i = 0;

for(; i < length; i++)
{
if(inorder[preIn + i] == rootValue)
{
break;
}
}

if(i == length)
{
return NULL;
}

if(i > 0)
{
root->left = buildTreeCore(preorder, preStart + 1, inorder, preIn, i);
}

if(i < length - 1)
{
root->right = buildTreeCore(preorder, preStart + i + 1, inorder, preIn + i + 1, length - 1 - i);
}

return root;
}