二叉树的前序、中序、后序遍历（python递归）

先序遍历

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)