Welcome to Subscribe On Youtube
Question
Formatted question description: https://leetcode.ca/all/222.html
222 Count Complete Tree Nodes
Given a complete binary tree, count the number of nodes.
Note:
Definition of a complete binary tree from Wikipedia:
In a complete binary tree
every level, except possibly the last, is completely filled,
and all nodes in the last level are as far left as possible.
It can have between 1 and 2^h nodes inclusive at the last level h.
Example:
Input:
1
/ \
2 3
/ \ /
4 5 6
Output: 6
@tag-tree
Algorithm
First, we need a getHeight
function, which is used to count the maximum height of the left subtree of the current node.
- Call the
getHeight
function on the current node to get the maximum height h of the left subtree. If returned is -1, it means that the current node does not exist and returns 0 directly. - Otherwise, call the
getHeight
function on theright
child.- If the return value is
h-1
, it means that the left sub-tree is a perfect binary tree, and the number of nodes in the left sub-tree is2^h-1
. Adding the current node, there are2^h
in total, that is,1<<h
,- then, add the return value of calling the recursive function on the
right
child node.
- then, add the return value of calling the recursive function on the
- If the return value of calling the
getHeight
function on the right subnode is noth-1
, it means that the right subtree must be a perfect tree, and the height ish-1
,- then the number of summary points is
2^(h-1) -1
, plus the current node is2^(h-1)
, that is,1<<(h-1)
, and then add the return value of calling the recursive function on theleft
child node
- then the number of summary points is
- If the return value is
Code
Java
-
public class Count_Complete_Tree_Nodes { /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution_optimize { public int countNodes(TreeNode root) { int res = 0; int h = getHeight(root); // here h is actually h-1 if (h < 0) return 0; if (getHeight(root.right) == h - 1) { return (1 << h) + countNodes(root.right); } else { // getHeight(root.right) == h - 2 return (1 << (h - 1)) + countNodes(root.left); } } int getHeight(TreeNode node) { return node != null ? (1 + getHeight(node.left)) : -1; } } public class Solution { public int countNodes(TreeNode root) { if (root == null) { return 0; } int leftHeight = findLeftHeight(root); int rightHeight = findRightHeight(root); if (leftHeight == rightHeight) { return (2 << (leftHeight - 1)) - 1; } return 1 + countNodes(root.left) + countNodes(root.right); } private int findLeftHeight(TreeNode root) { if (root == null) { return 0; } int height = 1; while (root.left != null) { height++; root = root.left; } return height; } private int findRightHeight(TreeNode root) { if (root == null) { return 0; } int height = 1; while (root.right != null) { height++; root = root.right; } return height; } } }
-
// OJ: https://leetcode.com/problems/count-complete-tree-nodes/ // Time: O(H^2) // Space: O(H) class Solution { int countLeft(TreeNode *root) { int cnt = 0; for (; root; ++cnt, root = root->left); return cnt; } int countRight(TreeNode *root) { int cnt = 0; for (; root; ++cnt, root = root->right); return cnt; } public: int countNodes(TreeNode* root) { if (!root) return 0; int left = countLeft(root), right = countRight(root); if (left == right) return (1 << left) - 1; return countNodes(root->left) + countNodes(root->right) + 1; } };
-
# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def countNodes(self, root: Optional[TreeNode]) -> int: def depth(root): d = 0 while root: d += 1 root = root.left return d if root is None: return 0 left, right = depth(root.left), depth(root.right) if left == right: return (1 << left) + self.countNodes(root.right) return (1 << right) + self.countNodes(root.left) ############ ''' >>> 2 ** 3 8 >>> 3 ** 2 9 ''' # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def getHeight(self, root): height = 0 while root: height += 1 root = root.left return height def countNodes(self, root): count = 0 while root: l, r = map(self.getHeight, (root.left, root.right)) if l == r: count += 2 ** l root = root.right else: count += 2 ** r root = root.left return count