二叉树的几个基础遍历算法代码

最新推荐文章于 2024-09-27 11:10:57 发布

liujinliu001

最新推荐文章于 2024-09-27 11:10:57 发布

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分类专栏：算法文章标签：二叉树遍历算法

本文链接：https://blog.youkuaiyun.com/github_25679381/article/details/50634405

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本文介绍了二叉树的先序、中序和后序遍历算法，并提供了详细的Python实现代码。通过这些算法，读者可以更好地理解二叉树的数据结构及其操作。

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先序遍历

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

class Solution:
    # @param root, a tree node
    # @return a list of integers
    def preorderTraversal(self, root):
        stack = []
        ret = []
        while root or len(stack):
            if root:
                stack.append(root)
                ret.append(root.val)
                root = root.left
            else:
                root = stack.pop()
                root = root.right
        return ret

后序遍历

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

class Solution:
    # @param root, a tree node
    # @return a list of integers
    def postorderTraversal(self, root):
            s=[]
            ret=[]
            s.append(root)
            curTree=None
            preTree=None
            while 0 != len(s):
                curTree=s[-1]
                if None == curTree:
                    return ret
                if( (None==curTree.left and None==curTree.right) or 
                (None != preTree and (preTree==curTree.left or preTree==curTree.right)) ):
                    ret.append(curTree.val)
                    s.pop()
                    preTree=curTree
                else:
                    if None != curTree.right:
                        s.append(curTree.right)
                    if None != curTree.left:
                        s.append(curTree.left)

            return ret

中序遍历

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

class Solution:
    # @param root, a tree node
    # @return a list of integers
    def inorderTraversal(self, root):
        stack=[]
        ret = []
        while root or len(stack):
            if root:
                stack.append(root)
                root=root.left
            else:
                tmp = stack.pop()
                root = tmp.right
                ret.append(tmp.val)
        return ret