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;
    }

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