从前序遍历迭代（非递归）构建二叉搜索树

Question

The following is the code to converted a preorder traversal of a Binary Search Tree to the original tree.

The following code takes an array of integers, which represent the pre order traversal of aa Binary search tree. The root of the construct tree is returned.

struct Node* constructTree(int pre[], int size)
{
    stack<struct Node* > s;
    int i;
    struct Node* root=newNode(pre[0]);
    struct Node* temp;
    struct Node* top_node;
    s.push(root);
    for(i=1;i<size;i++)
    {
        temp=NULL;
        while(!s.empty()&&pre[i]>(s.top()->data))
            {
                temp=s.top();
                s.pop();
            }
            if(temp==NULL)
            {
                top_node=s.top();
                top_node->left=newNode(pre[i]);
                s.push(top_node->left);
            }else
            {
                temp->right=newNode(pre[i]);
                s.push(temp->right);
            }
    }


    return root;
}

Source: http://www.geeksforgeeks.org/construct-bst-from-given-preorder-traversal-set-2/

I have trouble understanding this code. Can anybody help me understand the following:

1. At any given iteration, what values are stored in the stack, in relation to the current value being pointed out by pre[i]

2. Is there any other iterative method for constructing a BST from a given preorder traversal?

Thank you.

Answer 1

在构造包含pre[i]的节点的迭代之后，堆栈包含顶部的节点，在该节点下，它的最叶到最根的祖先只具有一个孩子，从上到下存储。

Answer 2

Check if this works:

public:
TreeNode* bstFromPreorder(vector<int>& preorder) {
    TreeNode *root = new TreeNode(preorder[0]);
    stack<TreeNode*> nodes;
    nodes.push(root);
    for (int i = 1; i < preorder.size(); i++) {
        TreeNode *temp = new TreeNode(preorder[i]);
        if (temp->val < nodes.top()->val)
            nodes.top()->left = temp;
        else {
            TreeNode *prev;
            while (!nodes.empty() && nodes.top()->val < temp->val) {
                prev = nodes.top();
                nodes.pop();
            }
            prev->right = temp;
        }
        nodes.push(temp);
    }
    return root;
}