def insert(self, v: int) -> int:
     n = len(self.memo)
     p_node = self.memo[(n - 1) // 2]
     if n % 2 == 1:  # adds left child
         p_node.left = TreeNode(v)
         self.memo.append(p_node.left)
     else:  # adds right child
         p_node.right = TreeNode(v)
         self.memo.append(p_node.right)
     return p_node.val
示例#2
0
 def invert_tree(self, root: TreeNode) -> TreeNode:
     '''
         翻转二叉树
     Args:
         root: 根节点
     Returns:
         TreeNode
     '''
     if root == None:
         return None
     left = self.invert_tree(root.left)
     right = self.invert_tree(root.right)
     root.left = right
     root.right = left
     return root
示例#3
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 def deserialize(self, data: str) -> TreeNode:
     if not data or data == '[]': return None
     data = data[1:-1].split(',')
     root = TreeNode(int(data[0]))
     idx, queue = 0, [root]
     while queue:
         cur = queue.pop(0)
         if idx * 2 + 1 < len(data) and data[idx * 2 + 1] != 'null':
             cur.left = TreeNode(int(data[idx * 2 + 1]))
             queue.append(cur.left)
         if idx * 2 + 2 < len(data) and data[idx * 2 + 2] != 'null':
             cur.right = TreeNode(int(data[idx * 2 + 2]))
             queue.append(cur.right)
         idx += 1
     return root
 def flatten_for_list(self, root: TreeNode, current_node) -> TreeNode:
     '''
         循环递归树
     Args:
         root: 根节点
         current_node: 当前节点
     Returns:
         root
     '''
     if not root:
         return current_node
     current_node = self.flatten_for_list(root.right, current_node)
     current_node = self.flatten_for_list(root.left, current_node)
     root.left = None
     root.right = current_node
     return root
示例#5
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def merge_trees(t1: TreeNode, t2: TreeNode) -> TreeNode:
    '''
        合并两棵二叉树
    Args:
        t1: 二叉树t1
        t2: 二叉树t2
    Returns:
        合并后的二叉树
    '''
    if t1 == None:
        return t2
    if t2 == None:
        return t1
    t1.val += t2.val
    t1.left = merge_trees(t1.left, t2.left)
    t1.right = merge_trees(t1.right, t2.right)
    return t1
示例#6
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    def insert(tree: TreeNode, data):
        if tree is None:
            return TreeNode(data)

        p = tree
        if data > p.val:
            if p.right is None:
                p.right = TreeNode(data)
                return tree
            else:
                return BinarySearchTree.insert(p.right, data)
        else:  # data <= p.val
            if p.left is None:
                p.left = TreeNode(data)
                return tree
            else:
                return BinarySearchTree.insert(p.left, data)
示例#7
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    def insert_iterative(tree: TreeNode, data):
        if tree is None:
            return TreeNode(data)

        p = tree
        while p is not None:
            if data > p.val:
                if p.right is None:
                    p.right = TreeNode(data)
                    return tree
                else:
                    p = p.right
            else:  # data <= p.val
                if p.left is None:
                    p.left = TreeNode(data)
                    return tree
                else:
                    p = p.left
 def convert_bst(self, root: TreeNode) -> TreeNode:
     '''
         二分查找树
     Args:
         root: 根节点
     Returns:
         节点的值
     '''
     global total_sum
     if not root:
         self.convert_bst(root.right)
         total_sum += root.val
         root.val = total_sum
         self.convert_bst(root.left)
     return root
示例#9
0

def max_depth(root: TreeNode):
    if not root: return 0
    q = [root]
    step = 1
    all_step = []
    while q:
        n = len(q)
        for _ in range(n):
            node = q.pop(0)
            if not node:
                continue
            if not node.left and not node.right:
                all_step.append(step)

            q.append(node.left)
            q.append(node.right)
        step += 1
    return max(all_step)


t = TreeNode(3)
t.left = TreeNode(9)

t.right = TreeNode(20)
t.right.left = TreeNode(15)
t.right.right = TreeNode(7)

print(min_depth(t))
print(max_depth(t))
            root: 根节点
        '''
        if root.left:
            self.helper(result, root.left)
        if root.right:
            self.helper(result, root.right)
        result.append(root.val)

    def post_order_traversal(self, root: TreeNode) -> List[int]:
        '''
            后序遍历
        Args:
            res: 链表
            root: 根节点
        '''
        result = []
        if not root:
            return result
        self.helper(result, root)
        return result


if __name__ == '__main__':
    arr = [1, 2, 3, 4, 5]
    tree_node = TreeNode()
    tree = tree_node.create_binary_tree_array(arr)
    solution = Solution()
    result = solution.post_order_traversal(tree)
    print(result)
    assert result[len(result) - 1] == 1
示例#11
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        Returns:
            布尔值
        '''
        queue = []
        queue.append(root)
        queue.append(root)
        while queue:
            r1 = queue.pop()
            r2 = queue.pop()
            if not r1 and not r2:
                continue
            if not r1 or not r2:
                return False
            if r1.val != r2.val:
                return False
            queue.append(r1.left)
            queue.append(r2.right)
            queue.append(r1.right)
            queue.append(r2.left)
        return True


if __name__ == '__main__':
    arr1 = [1, 2, 2]
    tree_node = TreeNode()
    root = tree_node.create_binary_tree_array(arr1)
    solution = Solution()
    result = solution.is_symmetric2(root)
    print(result)
    assert result == True
示例#12
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 def build(self, array, p, q):
     nodes = {}
     for v in array:
         if v != None:
             nodes[v] = TreeNode(v)
     return nodes, nodes[p], nodes[q]
示例#13
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Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.

"""
from tree.tree_node import TreeNode


def invert_binary_tree(root: TreeNode):
    if not root:
        return None
    stack = [root]  # stack存放还没有做对换的节点
    while stack:
        node = stack.pop()
        if not node:
            continue
        node.left, node.right = node.right, node.left
        stack.append(node.left)
        stack.append(node.right)
    return root


t = TreeNode(4)
t.left = TreeNode(2)
t.left.left = TreeNode(1)
t.left.right = TreeNode(3)

t.right = TreeNode(7)
t.right.left = TreeNode(6)
t.right.right = TreeNode(9)

print(invert_binary_tree(t))
示例#14
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    '''
    if t1 == None:
        return t2
    if t2 == None:
        return t1
    stack = []
    stack.append((t1, t2))
    while stack:
        t = stack.pop()
        if t[0] == None or t[1] == None:
            continue
        t[0].val += t[1].val
        if t[0].left == None:
            t[0].left = t[1].left
        else:
            stack.append((t[0].left, t[1].left))
        if t[0].right == None:
            t[0].right = t[1].right
        else:
            stack.append((t[0].right, t[1].right))
    return t1


if __name__ == '__main__':
    tree_node = TreeNode()
    t1 = tree_node.create_binary_tree_array([1, 2, 3], 0)
    t2 = tree_node.create_binary_tree_array([1, 2, 3], 0)
    t = merge_trees2(t1, t2)
    t_arr = per_order_traversal(t)
    print(t_arr)
示例#15
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        root: 节点root
        p: 节点p
        q: 节点q
    Returns:
        最小公共祖先
    '''
    if not root:
        return root
    if not p or not q:
        return root
    left = lowest_common_ancestor(root.left, p, q)
    right = lowest_common_ancestor(root.right, p, q)
    if not left and not right:
        return None
    if left and right:
        return root
    if not left:
        return right
    else:
        return left


if __name__ == '__main__':
    nums = [3, 5, 1, 6, 2, 0, 8, None, None, 7, 4]
    tree_node = TreeNode()
    p = TreeNode(5)
    q = TreeNode(1)
    root = tree_node.create_binary_tree_array(nums)
    result = lowest_common_ancestor(root, p, q)
    print(result)
示例#16
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    @staticmethod
    def height(root: TreeNode):
        if root is None:
            return 0
        else:
            # compute the height of each subtree
            l_height = BinaryTreeTraversal.height(root.left)
            r_height = BinaryTreeTraversal.height(root.right)

            # return the larger one
            return max(l_height, r_height) + 1


if __name__ == '__main__':
    # construct a binary tree
    d = TreeNode('d')
    e = TreeNode('e')
    f = TreeNode('f')
    g = TreeNode('g')

    b = TreeNode('b', d, e)
    c = TreeNode('c', f, g)

    a = TreeNode('a', b, c)

    # traverse the binary via different approaches
    print('pre-order:', end='\t\t')
    BinaryTreeTraversal.pre_order(a)

    print('\nin-order:', end='\t\t')
    BinaryTreeTraversal.in_order(a)
        self.in_order(root.left)

        if self.pre and self.pre.val > root.val:
            if not self.first: self.first = self.pre
            self.second = root

        self.pre = root

        self.in_order(root.right)

    def recover_tree(self, root: TreeNode):

        self.in_order(root)

        self.first.val, self.second.val = self.second.val, self.first.val


r = TreeNode(3)
r.left = TreeNode(1)
r.right = TreeNode(4)
r.right.left = TreeNode(2)

recover = RecoverTree()
recover.recover_tree(r)






示例#18
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    def invert_tree2(self, root: TreeNode) -> TreeNode:
        '''
            翻转二叉树
        Args:
            root: 根节点
        Returns:
            TreeNode
        '''
        if root == None:
            return None
        node_list = []
        node_list.append(root)
        while len(node_list) > 0:
            cur = node_list.pop()
            if cur.left:
                node_list.append(cur.left)
            if cur.right:
                node_list.append(cur.right)
            temp = cur.right
            cur.right = cur.left
            cur.left = temp
        return root


if __name__ == '__main__':
    arr = [4, 2, 7, 1, 3, 6, 9]
    root = TreeNode.create_binary_tree_array(arr)
    solution = Solution()
    result = solution.invert_tree(root)
    assert 4 == result.val
示例#19
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        elif p.right is not None:
            child = p.right

        if pp is None:
            tree = child
        elif pp.left == p:
            pp.left = child
        else:
            pp.right = child

        return tree


if __name__ == '__main__':
    # construct a binary tree
    m_19 = TreeNode(19)
    n_27 = TreeNode(27)
    o_55 = TreeNode(55)

    h_15 = TreeNode(15)
    i_17 = TreeNode(17)
    j_25 = TreeNode(25, m_19, n_27)
    k_51 = TreeNode(51, right=o_55)
    l_66 = TreeNode(66)

    d_13 = TreeNode(13, right=h_15)
    e_18 = TreeNode(18, i_17, j_25)
    f_34 = TreeNode(34)
    g_58 = TreeNode(58, k_51, l_66)

    b_16 = TreeNode(16, d_13, e_18)
    def is_valid_BST_2(self, root: TreeNode) -> bool:
        '''
            中序遍历二叉树判断是不是BST树
        Args:
            root: 根节点
        Returns:
            布尔值
        '''
        pre = -sys.maxsize
        stack = []
        while not root or len(stack) < 1:
            while not root.left:
                stack.append(root.left)
                root = root.left
            # 栈退出一个元素
            root = stack.pop()
            if root.val <= pre:
                return False
            pre = root.val
            root = root.right
        return True


if __name__ == '__main__':
    nums = [2, 1, 3]
    tree = TreeNode()
    root = tree.create_binary_tree_array(nums)
    solution = Solution()
    result = solution.is_valid_BST(root)
    assert result == True