Leetcode学习笔记：#1008. Construct Binary Search Tree from Preorder Traversal

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.)

实现：

int i = 0;
public TreeNode bstFromPreorder(int[] A){
	return bstFromPreorder(A, Integer.MAX_VALUE);
}

public TreeNode bstFromPreorder(int[] A, int bound){
	if(i == A.length || A[i] > bound) 
		return null;
	TreeNode root = new TreeNode(A[i++]);
	root.left = bstFromPreorder(A, root.val);
	root.right = bstFromPreorder(A, bound);
	return root;
}	

思路：
前序遍历，递归。前序遍历说明第一个数字必定为根节点，则只要把数组第一个数字设为根节点，再递归得出左子树和右子树的根节点即可。