为什么这个 function 检查二叉树平衡的时间复杂度是 O(n log n)？

Question

Prompt from Cracking the Coding Interview by Gayle Laakmann McDowell:

Implement a function to check if a binary tree is balanced. For this problem, a balanced tree is defined such that the heights of the two subtrees of any node do not differ by more than one.

(Example implementation below.)

Question: Can you help me understand why the author states that isBalanced has a time complexity of O(n log n) ? I get it to some degree and can memorize this just fine, but I can't conceptualize why this is the case like I can for other time complexities like O(n^2) .

int getHeight(TreeNode root) {
  if (root == null) { return -1; }
  return Math.max(getHeight(root.left), getHeight(root.right)) + 1;
}

boolean isBalanced(TreeNode root) {
  if (root == null) { return true; }

  int heightDiff = getHeight(root.left) - getHeight(root.right);
  if (Maths.abs(heightDiff) > 1) {
    return false;
  } else {
    return isBalanced(root.left) && isBalanced(root.right);
  }
}

// isBalanced([some node]) --> true/false

How do I visualize why isBalanced is considered O(n log n) ?

Answer 1

Lets say you have BST

      4        // No of nodes 1
    /    \
   2      6    // No of nodes 2
 /  \    /  \
1    3  5    7 // No of nodes 4

Your function isBalanced is going through all nodes including the ones with no children and call getHeight to calculate the height of left and right child.

The recursive relation of the function would be

which is derived from Master Theorem

a = number of sub-problems in recursion

n/b = size of each sub-problem

f(n) = cost of work that has to be done outside the recursive calls

a is 2 because we have to visit both children nodes of each parent

b is 2 because if you notice the number of nodes reduce by half each time you go a level above (from bottom).

f(n) is n because we have to call getHeight() on each node

Which satisfies the second case of Master Theorem that is

Putting the values to prove f(n) = O(n^logb(a))

Thus we get the time complexity of O(n log n)