Question

我正在阅读以下数据，作为分配到二叉树（不是严格二叉搜索树）的一部分：

他们被读入python self.key，self.left和self.right中的三个列表，其中第一行的整数n是节点数。接下来的n行是键，左，右。左边是父亲的左子女的密钥是key[left]，同样右子的密钥是key[right]，所以例如第一行是4的关键是根和{{ 1}}含义2是4的左子，key[1]意味着5是4的右子，依此类推，左和右的-1表示该键是叶子：

Tree structure for this example

问题是正在添加根的左右子节点，但是没有添加这些节点的子节点。我是否正确地向树中添加节点？我不能只根据密钥的值添加它们，因为它不是严格的二进制搜索树，因为其他一些例子清楚，例如 root = 0 和 left child = 70 和 right child = 20 。 inOrder遍历的输出为key[2]（应为2 4 5），这使我相信我没有添加其他节点。任何有关添加方法的帮助都将受到赞赏......

1 2 3 4 5

Answer 1

谢谢我让它工作 - 我将节点Node(key[i])添加到字典和self.nodes[val] = [node, node.l, node.r]，并在添加左边并递归搜索字典以获取inOrder，preOrder和postOrder树遍历。

 class Node:
    def __init__(self, val):
        self.l = None
        self.r = None
        self.v = val

class Tree:

    def __init__(self):
        self.root = None
        self.nodes = {}

    def getRoot(self):
        return self.root

    def add_root(self, val):
        if(self.root is None):
            self.root = Node(val)
            self.nodes[val] = [self.root,-1,-1]

    def add_left(self, val, node):
        if(node.l is None):
            node.l = Node(val)
            self.nodes[node.v][1] = node.l

    def add_right(self, val, node):
        if(node.r is None):
            node.r = Node(val)
            self.nodes[node.v][2] = node.r

    def inOrder(self):
        self.result = []
        if(self.root is not None):
            self._inOrder(self.root, self.result)
            return self.result
        else:
            print('root is None')

    def _inOrder(self, node, result):
        if(node is not None):
            try:
                self._inOrder(self.nodes[node.v][1], self.result)
            except (IndexError, AttributeError):
                pass
            self.result.append(node.v)
            try:
                self._inOrder(self.nodes[node.v][2], self.result)
            except (IndexError, AttributeError):
                pass
    def preOrder(self):
        self.result = []
        if(self.root is not None):
            self._preOrder(self.root, self.result)
            return self.result
        else:
            print('root is None')

    def _preOrder(self, node, result):
        if(node is not None):
            self.result.append(node.v)
            try:
                self._preOrder(self.nodes[node.v][1], self.result)
            except (IndexError, AttributeError):
                pass
            try:
                self._preOrder(self.nodes[node.v][2], self.result)
            except (IndexError, AttributeError):
                pass

    def postOrder(self):
        self.result = []
        if(self.root is not None):
            self._postOrder(self.root, self.result)
            return self.result
        else:
            print('root is None')

    def _postOrder(self, node, result):
        if(node is not None):
            try:
                self._postOrder(self.nodes[node.v][1], self.result)
            except (IndexError, AttributeError):
                pass
            try:
                self._postOrder(self.nodes[node.v][2], self.result)
            except (IndexError, AttributeError):
                pass
            self.result.append(node.v)

    def read(self):
        self.n = int(sys.stdin.readline())
        self.key = [0 for i in range(self.n)]
        self.left = [0 for i in range(self.n)]
        self.right = [0 for i in range(self.n)]
        for i in range(self.n):
            [a, b, c] = map(int, sys.stdin.readline().split())
            self.key[i] = a
            self.left[i] = b
            self.right[i] = c

        #adding root
        self.add_root(self.key[0])

        for i in range(1, self.n):
            self.nodes[self.key[i]] = [Node(self.key[i]),-1,-1]

        for i in range(0, self.n):
            if self.left[i] != -1:
                # adding the other left nodes
                self.add_left(self.key[self.left[i]], self.nodes[self.key[i]][0])

            if self.right[i] != -1:
                # adding the other right nodes
                self.add_right(self.key[self.right[i]], self.nodes[self.key[i]][0])

二叉树添加左侧并添加右侧节点而不添加

1 个答案: