二叉树遍历python3代码（先序、中序、后序、层次）（递归、非递归）

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

 

（一）二叉树的中序遍历

递归：

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res=[]
        if root:
            res+=self.inorderTraversal(root.left)
            res.append(root.val)
            res+=self.inorderTraversal(root.right)
        return res

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

注：

1. 类中方法的自我调用

2. Python中list可以直接相加得到新的list：

ls1 = [1,2,3]
ls2 = [4,5,6]
print(ls1+ls2)

迭代：

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        # 迭代解法
        p = root
        res = []
        stack = []
        
        while p or stack:
            if p:
                stack.append(p)
                p = p.left
            else:
                tmp = stack.pop()
                res.append(tmp.val);
                p = tmp.right
        
        return res
                

（二）二叉树的先序（前序）遍历

递归：

class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        '''递归解法'''
        p =root
        res = []
        
        if p!=None:
            res.append(p.val)
            res += self.preorderTraversal(p.left)
            res += self.preorderTraversal(p.right)
            
        return res

迭代：

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

（三）二叉树的后序遍历

递归：

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        p = root
        res = []
        
        if p:
            res += self.postorderTraversal(p.left)
            res += self.postorderTraversal(p.right)
            res.append(p.val)
        return res

 

后序遍历参考资料

已有详细解释说明，不再说明。

迭代1：

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        '''先序遍历思想实现后续遍历'''
        p = root
        #res = []
        stack = []
        stack2 = []
        
        while p or stack:
            if p:
                stack2.append(p.val)
                stack.append(p)
                p = p.right
            else:
                temp = stack.pop()
                p = temp.left
        return stack2[::-1]
        

迭代2：

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        '''后序遍历双指针迭代算法'''
        if not root:   # 需要判断是否为空
            return []
        
        stack = []
        res = []
        prev = None
        curr = None
        stack.append(root)
        
        while stack:
            curr = stack[-1]
            if prev==None or prev.left==curr or prev.right==curr:
                if curr.left!=None:
                    stack.append(curr.left)
                elif curr.right!=None:
                    stack.append(curr.right)
            elif prev == curr.left:
                if curr.right!=None:
                     stack.append(curr.right)
            else:
                res.append(curr.val)
                stack.pop()         # 需要弹出
            prev = curr
        return res
                    

（四）二叉树的层次遍历

采用队列组织结构

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        
        queue = []
        res = []
        
        p = root
        queue.append(p)
        
        while queue:
            temp = queue.pop(0)
            res.append(temp.val)
            if temp.left!=None:
                queue.append(temp.left)
            if temp.right!=None:
                queue.append(temp.right)
        return res