【leetcode】1008. Construct Binary Search Tree from Preorder Traversal

題目以下：

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.

解題思路：以用例的輸入[8,5,1,7,10,12]爲例，很顯然8是根節點，8的左子樹有[5,1,7]，右子樹右[10,12]，左右子樹的分割點是後面第一個比根節點大的數。接下來再分別對[5,1,7]和[10,12]作一樣的操做，能夠知道5是8的左子樹根節點，1和7分別在其左右；而10是8的右子樹根節點，12爲右子樹節點。很顯然這是一個遞歸的過程，只要找到每一個子樹的根節點將其左右子樹劃分便可。

代碼以下：

# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None

class Solution(object): def build(self,node,preorder): if len(preorder) == 0: return left = [] for i in range(len(preorder)): if node.val < preorder[i]: break
            else: left.append(preorder[i]) right = preorder[len(left):] if len(left) >= 1: node.left = TreeNode(left.pop(0)) self.build(node.left,left) if len(right) >= 1: node.right = TreeNode(right.pop(0)) self.build(node.right,right) def bstFromPreorder(self, preorder): """ :type preorder: List[int] :rtype: TreeNode """ root = TreeNode(preorder.pop(0)) self.build(root,preorder) return root