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day-20.cpp
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day-20.cpp
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/*
Question: 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]
*/
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, int low, int high) {
if (low > high) return nullptr;
TreeNode* root = new TreeNode(preorder[low]);
if (low == high) return root;
int idx = low + 1;
while (idx <= high && preorder[idx] < preorder[low]) idx += 1;
root->left = buildTree(preorder, low + 1, idx - 1);
root->right = buildTree(preorder, idx, high);
return root;
}
TreeNode* bstFromPreorder(vector<int>& preorder) {
return buildTree(preorder, 0, preorder.size() - 1);
}
};