5月挑战Day24-Construct Binary Search Tree from Preorder Traversal(Medium)

Day24-Construct Binary Search Tree from Preorder Traversal(Medium)

问题描述：

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.)

It’s guaranteed that for the given test cases there is always possible to find a binary search tree with the given requirements.

依据前序遍历构造二叉搜索树

解法：

这道题好像四月挑战的时候就做过了。。。

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

class Solution:
    def bstFromPreorder(self, preorder: List[int]) -> TreeNode:
        if len(preorder) == 0:
            return None
        #因为先序遍历所以第一个元素一定为跟节点。
        root = TreeNode(preorder[0])
        #先序遍历先访问根节点然后左子树，然后右子树，同时左子树的节点一定比根节点的数小，右子树的节点一定比根节点的数大。所以向后寻找第一个比根节点大的数字就是右子树的根节点。然后递归求解左右子树即可
        i = 1
        while i < len(preorder):
            if preorder[i] > root.val:
                break
            i += 1
        root.left = self.bstFromPreorder(preorder[1:i])
        root.right = self.bstFromPreorder(preorder[i:])
        return root

解法二：

虽然上个月刚做过，但是今天不写道算法题就过去，实在不太舒服啊（受虐狂。。）于是上discuss中找了另一种解法，同时感叹leetcode真是能人辈出，各种新奇的解法都有，这里贴上一个我认为还是很大众的解法出来。

就是不用递归而用栈，我们用栈存储左子树上的节点，遇到右子树上的点时，我们就从栈中往外推值，直到找到右子树的根节点为止。然后将右子树再送入栈中循环。

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

class Solution:
    def bstFromPreorder(self, preorder: List[int]) -> TreeNode:
        if len(preorder) == 0:
            return None
        ##换一种写法，不使用递归求解而是用栈。
        root = TreeNode(preorder[0])
        help_stack = [root]
        for val in preorder[1:]:
            if help_stack[-1].val > val:
                help_stack[-1].left = TreeNode(val)
                help_stack.append(help_stack[-1].left)
                
            else:
                while help_stack and help_stack[-1].val < val:
                    temp = help_stack.pop()
                temp.right = TreeNode(val)
                help_stack.append(temp.right)
        return root