25期代码随想录算法训练营第十四天 | 二叉树 | 递归遍历、迭代遍历

递归遍历

前序遍历

python 复制代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []

        left = self.preorderTraversal(root.left)
        right = self.preorderTraversal(root.right)

        return [root.val] + left + right

中序遍历

python 复制代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []

        left = self.inorderTraversal(root.left)
        right = self.inorderTraversal(root.right)

        return left + [root.val] + right

后序遍历

python 复制代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []

        left = self.postorderTraversal(root.left)
        right = self.postorderTraversal(root.right)

        return left + right + [root.val]

迭代遍历

前序遍历

python 复制代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []

        stack = [root]
        res = []

        while stack:
            cur = stack.pop()
            res.append(cur.val)  # 在这里加入节点值

            # 先右后左地加入子节点到栈中
            if cur.right:
                stack.append(cur.right)
            if cur.left:
                stack.append(cur.left)

        return res

中序遍历

为什么这样能实现左中右的逻辑？

为什么需要使用while cur or stack?

python 复制代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []

        stack = []
        res = []
        cur = root

        while cur or stack:
            if cur:
                stack.append(cur)
                cur = cur.left
            else:
                cur = stack.pop()
                res.append(cur.val)
                cur = cur.right

        return res

后序遍历

中右左 ---->再颠倒顺序成为左右中

python 复制代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []

        res = []
        stack = [root]

        while stack:
            node = stack.pop()
            res.append(node.val)

            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)
        
        return res[::-1]

25期代码随想录算法训练营第十四天 | 二叉树 | 递归遍历、迭代遍历

目录

递归遍历

前序遍历

中序遍历

后序遍历

迭代遍历

前序遍历

中序遍历

后序遍历