資訊專(zhuān)欄INFORMATION COLUMN
摘要：題目此題和很類(lèi)似，所以第一反應(yīng)使用遞歸做。遞歸的解法過(guò)不了，會(huì)顯示超時(shí)比遞歸的更好的方法是用，比較難想到，代碼參考了思路是用一個(gè)單調(diào)遞減棧尋找最大值。這個(gè)操作可以幫我們順利找到左子樹(shù)和父節(jié)點(diǎn)。
題目
Given an integer array with no duplicates. A max tree building on this array is defined as follow:
The root is the maximum number in the array
The left subtree and right subtree are the max trees of the subarray divided by the root number.
Construct the max tree by the given array.
此題和Construct Binary Tree from Preorder and Inorder Traversal 很類(lèi)似， 所以第一反應(yīng)使用遞歸做。遞歸的解法過(guò)不了lintcode，會(huì)顯示超時(shí)：
class Solution:
"""
@param: A: Given an integer array with no duplicates.
@return: The root of max tree.
"""
def maxTree(self, A):
def helper(A, start, end):
if start > end:
return None
elif start == end:
return TreeNode(A[start])
maxVal = 0
maxIndex = -1
for i in range(start, end+1):
if A[i] > maxVal:
maxVal, maxIndex = A[i], i
root = TreeNode(maxVal)
root.left = helper(A, start, maxIndex - 1)
root.right = helper(A, maxIndex + 1, end)
return root
if A is None or len(A) == 0:
return None
return helper(A, 0, len(A)-1)
比遞歸的更好的方法是用stack，比較難想到，代碼參考了：https://lefttree.gitbooks.io/...
思路是用一個(gè)單調(diào)遞減棧尋找最大值。每遍歷到一個(gè)新的數(shù)字A[i]，我們就把比這個(gè)數(shù)字小的都彈棧，直到剩下比此數(shù)字大的數(shù)字。這個(gè)操作可以幫我們順利找到左子樹(shù)和父節(jié)點(diǎn)。這個(gè)解法讓我想到了經(jīng)典的樹(shù)的非遞歸遍歷，需要再?gòu)?fù)習(xí)一下。
class Solution:
"""
@param: A: Given an integer array with no duplicates.
@return: The root of max tree.
"""
def maxTree(self, A):
stack = []
for val in A:
node = TreeNode(val)
while len(stack) > 0 and val > stack[-1].val:
node.left = stack.pop()
if len(stack) > 0:
stack[-1].right = node
stack.append(node)
return stack[0]
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