二叉树的遍历(python)

本文主要是介绍二叉树的遍历(python)，对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

一、中序

1. 递归

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        res = []
        res.extend(self.inorderTraversal(root.left))
        res.append(root.val)
        res.extend(self.inorderTraversal(root.right))
        return res

2. 迭代

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        p = root
        stack = []
        res = []
        while p or stack:
            while p:
                stack.append(p)
                p = p.left
            p = stack.pop()
            res.append(p.val)
            p = p.right
        return res

3. Morris

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        p = root
        while p:
            if p.left:
                predecessor = p.left
                while predecessor.right:
                    predecessor = predecessor.right
                predecessor.right = p
                temp = p
                p = p.left
                temp.left = None
            else:
                res.append(p.val)
                p= p.right
        return res

二、前序

1. 递归

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        res = [root.val]
        res.extend(self.inorderTraversal(root.left))
        res.extend(self.inorderTraversal(root.right))
        return res

2. 迭代

class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        stack = []
        p = root
        while stack or p:
            while p:
                res.append(p.val)
                stack.append(p.right)
                p = p.left
            p = stack.pop()
        return res

3. Morris

class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        res = list()
        p1 = root
        while p1:
            p2 = p1.left
            if p2:
                while p2.right and p2.right != p1:
                    p2 = p2.right
                if not p2.right:
                    res.append(p1.val)
                    p2.right = p1
                    p1 = p1.left
                    continue
                else:
                    p2.right = None
            else:
                res.append(p1.val)
            p1 = p1.right
        return res

三、后序

1. 递归

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        res = []
        res.extend(self.inorderTraversal(root.left))
        res.extend(self.inorderTraversal(root.right))
        res.append(root.val)
        return res

2. 迭代

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        stack = []
        p= root
        while stack or p:
            while p:
                res.append(p.val)
                stack.append(p.left)
                p= p.right
            p= stack.pop()
        res.reverse()
        return res

3. Morris

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        def addPath(node: TreeNode):
            count = 0
            while node:
                count += 1
                res.append(node.val)
                node = node.right
            i, j = len(res) - count, len(res) - 1
            while i < j:
                res[i], res[j] = res[j], res[i]
                i += 1
                j -= 1
        
        res = []
        p1 = root

        while p1:
            p2 = p1.left
            if p2:
                while p2.right and p2.right != p1:
                    p2 = p2.right
                if not p2.right:
                    p2.right = p1
                    p1 = p1.left
                    continue
                else:
                    p2.right = None
                    addPath(p1.left)
            p1 = p1.right
        
        addPath(root)
        return res

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