1、Binary Tree Preorder Traversal---leetcode144
#coding:utf-8
class Solution:
#根左右
def preorderTraversal(self, root):
if not root:
return []
return [root.val] + self.preorderTraversal(root.left) + self.preorderTraversal(root.right)
# 给定二叉树的前序遍历和中序遍历,获得该二叉树
def getBSTwithPreTin(self, pre, tin):
if len(pre)==0 | len(tin)==0:
return None
root = treeNode(pre[0])
for order,item in enumerate(tin):
if root .val == item:
root.left = self.getBSTwithPreTin(pre[1:order+1], tin[:order])
root.right = self.getBSTwithPreTin(pre[order+1:], tin[order+1:])
return root
class treeNode:
def __init__(self, x):
self.left = None
self.right = None
self.val = x
if __name__ == '__main__':
flag = "printTreeNode"
solution = Solution()
preorder_seq = [1, 2, 4, 7, 3, 5, 6, 8]
middleorder_seq = [4, 7, 2, 1, 5, 3, 8, 6]
treeRoot1 = solution.getBSTwithPreTin(preorder_seq, middleorder_seq)
if flag == "printTreeNode":
newArray = solution.preorderTraversal(treeRoot1)
print(newArray)
2、Binary Tree Inorder Traversal---leetcode94
#coding:utf-8
class Solution:
#左根右
def inorderTraversal(self, root):
if not root:
return []
return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right)
# 给定二叉树的前序遍历和中序遍历,获得该二叉树
def getBSTwithPreTin(self, pre, tin):
if len(pre)==0 | len(tin)==0:
return None
root = treeNode(pre[0])
for order,item in enumerate(tin):
if root .val == item:
root.left = self.getBSTwithPreTin(pre[1:order+1], tin[:order])
root.right = self.getBSTwithPreTin(pre[order+1:], tin[order+1:])
return root
class treeNode:
def __init__(self, x):
self.left = None
self.right = None
self.val = x
if __name__ == '__main__':
flag = "printTreeNode"
solution = Solution()
preorder_seq = [1, 2, 4, 7, 3, 5, 6, 8]
middleorder_seq = [4, 7, 2, 1, 5, 3, 8, 6]
treeRoot1 = solution.getBSTwithPreTin(preorder_seq, middleorder_seq)
if flag == "printTreeNode":
newArray = solution.inorderTraversal(treeRoot1)
print(newArray)
3、Binary Tree Posorder Traversal---leetcode145
#coding:utf-8
class Solution:
#左右根
def postorderTraversal(self, root):
if not root:
return []
return self.postorderTraversal(root.left) + self.postorderTraversal(root.right)+[root.val]
# 给定二叉树的前序遍历和中序遍历,获得该二叉树
def getBSTwithPreTin(self, pre, tin):
if len(pre)==0 | len(tin)==0:
return None
root = treeNode(pre[0])
for order,item in enumerate(tin):
if root .val == item:
root.left = self.getBSTwithPreTin(pre[1:order+1], tin[:order])
root.right = self.getBSTwithPreTin(pre[order+1:], tin[order+1:])
return root
class treeNode:
def __init__(self, x):
self.left = None
self.right = None
self.val = x
if __name__ == '__main__':
flag = "printTreeNode"
solution = Solution()
preorder_seq = [1, 2, 4, 7, 3, 5, 6, 8]
middleorder_seq = [4, 7, 2, 1, 5, 3, 8, 6]
treeRoot1 = solution.getBSTwithPreTin(preorder_seq, middleorder_seq)
if flag == "printTreeNode":
newArray = solution.postorderTraversal(treeRoot1)
print(newArray)