Construct Binary Search Tree from Preorder Traversal

Description

Return the root node of a binary search tree that matches the given preorder traversal.

(Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val. Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.)

Example 1:

Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]

Note:

1. 1 <= preorder.length <= 100

2. The values of preorder are distinct.

Solution

class Solution {
public:
    TreeNode *bstFromPreorder(vector<int> &preorder) {
        if (preorder.empty())
            return nullptr;
        TreeNode *node, *curr, *root = new TreeNode(preorder[0]);
        stack<TreeNode *> st;
        st.push(root);
        for (int i = 1; i < preorder.size(); ++i) {
            int val = preorder[i];
            if (val < st.top()->val) {
                node = new TreeNode(val);
                st.top()->left = node;
                st.push(node);
            } else {
                do {
                    curr = st.top();
                    st.pop();
                } while (!st.empty() && st.top()->val < val);
                curr->right = new TreeNode(val);
                st.push(curr->right);
            }
        }
        return root;
    }
};