Is a binary tree a balanced structural state?

A binary tree is not necessarily a balanced structural state by itself. A balanced binary tree is one in which the heights of the left and right subtrees differ by no more than 1 for each node. This property ensures efficient access, insertion, and deletion of elements.

If a binary tree is unbalanced, in the worst case it can turn into a structure similar to a linked list, which degrades the performance of operations.

An example of checking for balance in Go (simplified):

func isBalanced(root *TreeNode) bool {
    if root == nil {
        return true
    }
    leftHeight := height(root.Left)
    rightHeight := height(root.Right)
    if abs(leftHeight - rightHeight) > 1 {
        return false
    }
    return isBalanced(root.Left) && isBalanced(root.Right)
}

func height(node *TreeNode) int {
    if node == nil {
        return 0
    }
    leftHeight := height(node.Left)
    rightHeight := height(node.Right)
    if leftHeight > rightHeight {
        return leftHeight + 1
    }
    return rightHeight + 1
}

func abs(a int) int {
    if a < 0 {
        return -a
    }
    return a
}

Thus, a binary tree can be either balanced or unbalanced, depending on its structure and construction rules.