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描述
Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root.
It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.
A binary search tree is a binary tree where for every node, any descendant of Node.left has a value strictly less than Node.val, and any descendant of Node.right has a value strictly greater than Node.val.
A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left, then traverses Node.right.
Example 1:
Input: preorder = [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]
Example 2:
Input: preorder = [1,3]
Output: [1,null,3]
Note:
1 <= preorder.length <= 100
1 <= preorder[i] <= 10^8
All the values of preorder are unique.
解析
根据题意，就是给出了一个二叉树的前序遍历结果，然后要求根据前序遍历构造一棵树，并且返回树的根节点。至于二叉树的定义和前序遍历的定义，题目中已经解释了，这里不再赘述。
当时自己解题的时候思路很乱，一直没有细扣例子中的规律，所以一直头大，直到看到了高人的解法豁然开朗。仔细研究发现其实就是个找规律的题目。
前序遍历有一个明显的特点，第一个出现的元素肯定是属于根节点，从第二个元素开始后面的元素肯定是属于左右子树的，而左子树都比根节点小，右子树都比右子树大，前序遍历会先遍历完左子树再遍历右子树，所以从第二个元素开始往后找出大于根节点的元素索引 i ，这就是右子树的根节点 ，左子树的元素则在 preorder[1:i] 列表中，右子树的元素则在 preorder[i:] 列表中，递归执行这个过程得到的树即为结果。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def bstFromPreorder(self, preorder):
"""
:type preorder: List[int]
:rtype: TreeNode
"""
if not preorder: return None
root = TreeNode(preorder[0])
N = len(preorder)
i = 1
while i < N:
if preorder[i] > root.val:
break
i += 1
root.left = self.bstFromPreorder(preorder[1:i])
root.right = self.bstFromPreorder(preorder[i:])
return root
运行结果
Runtime: 25 ms, faster than 43.13% of Python online submissions for Construct Binary Search Tree from Preorder Traversal.
Memory Usage: 13.6 MB, less than 21.80% of Python online submissions for Construct Binary Search Tree from Preorder Traversal.
解析
遍历列表元素也可以实现，原理和上面类似。
- 首先用 preorder[0] 初始化 root 根结点，使用 stack 来找不同子树的根节点
- 从第二个元素开始遍历 preorder ，元素有两种情况： （1）当元素小于 stack[-1] 时候，那当前元素属于的 stack[-1] 左子树，并将当前的节点加入到 stack 中 （2）否则，当元素大于 stack[-1] ，此时不断 stack.pop() 找到适合它的根节点，找到的根节点的右子树就是当前元素 ，并将当前结点加入到 stack 中
- 遍历结束得到的 root 就是答案
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def bstFromPreorder(self, preorder):
"""
:type preorder: List[int]
:rtype: TreeNode
"""
root = TreeNode(preorder[0])
stack = [root]
for value in preorder[1:]:
if value < stack[-1].val:
left = TreeNode(value)
stack[-1].left = left
stack.append(left)
else:
while stack and stack[-1].val<value:
tmp = stack.pop()
right = TreeNode(value)
tmp.right = right
stack.append(right)
return root
运行结果
Runtime: 39 ms, faster than 5.69% of Python online submissions for Construct Binary Search Tree from Preorder Traversal.
Memory Usage: 13.7 MB, less than 21.80% of Python online submissions for Construct Binary Search Tree from Preorder Traversal.
原题链接：leetcode.com/problems/co…
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