【Python版】二叉树的遍历

最新推荐文章于 2024-10-21 08:57:50 发布

原创最新推荐文章于 2024-10-21 08:57:50 发布 · 317 阅读

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二叉树同时被 2 个专栏收录

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算法与数据结构

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本文详细介绍了如何利用前序遍历和中序遍历、中序遍历和后序遍历来构造二叉树的方法，通过递归算法实现，并提供了具体的Python代码实现。深入理解二叉树的构造过程对于数据结构的学习至关重要。

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前序遍历和中序遍历树构造二叉树

 
"""
Definition of TreeNode:
"""
 
 
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
 
 
class Solution:
    """
    @param inorder: A list of integers that inorder traversal of a tree
    @param postorder: A list of integers that postorder traversal of a tree
    @return: Root of a tree
    """
 
    def buildTree(self, preorder, inorder):
        # write your code here
        if not inorder:
            return None
        root = TreeNode(preorder[0])
        rootpos = inorder.index(preorder[0])  # 使用index(ｘ)获取list中的ｘ的下标
        root.left = self.buildTree(preorder[1: rootpos + 1], inorder[: rootpos])
        root.right = self.buildTree(preorder[rootpos + 1:], inorder[rootpos + 1:])
        return root
 
    def printBinaTree(self, root):
        print(root.val, end=" ")
        if root.left:
            self.printBinaTree(root.left)
        if root.right:
            self.printBinaTree(root.right)
 
if __name__ == '__main__':
    inorder = [1, 2, 3]
    preorder = [2, 1, 3]
    sol = Solution()
    root = sol.buildTree(preorder, inorder)
    sol.printBinaTree(root)
    # print(ans)

中序遍历和后序遍历树构造二叉树

 
"""
Definition of TreeNode:
"""
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
 
 
class Solution:
    """
    @param inorder: A list of integers that inorder traversal of a tree
    @param postorder: A list of integers that postorder traversal of a tree
    @return: Root of a tree
    """
 
    def buildTree(self, inorder, postorder):
        # write your code here
        if not postorder:
            return None
        root = TreeNode(postorder[-1])
        rootpos = inorder.index(postorder[-1])
        # print(rootpos, root.val)
        root.left = self.buildTree(inorder[:rootpos], postorder[:rootpos])
        # print(rootpos, len(postorder), postorder[-1])
        root.right = self.buildTree(inorder[rootpos + 1:], postorder[rootpos:-1])
        return root
 
    def printBinaTree(self, root):
        print(root.val, end=" ")
        if root.left:
            self.printBinaTree(root.left)
        if root.right:
            self.printBinaTree(root.right)
 
 
if __name__ == '__main__':
    inorder = [1, 2, 3]
    postorder = [1, 3, 2]
    sol = Solution()
    root = sol.buildTree(inorder, postorder)
    sol.printBinaTree(root)