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 2h nodes inclusive at the last level h.

Example:

Input: 
    1
   / 
  2   3
 /   /
4  5 6

Output: 6

 https://www.cnblogs.com/grandyang/p/4567827.html

https://www.cnblogs.com/yrbbest/p/4993469.html

Explanation

The height of a tree can be found by just going left. Let a single node tree have height 0. Find the height h of the whole tree. If the whole tree is empty, i.e., has height -1, there are 0 nodes.

Otherwise check whether the height of the right subtree is just one less than that of the whole tree, meaning left and right subtree have the same height.

• If yes, then the last node on the last tree row is in the right subtree and the left subtree is a full tree of height h-1. So we take the 2^h-1 nodes of the left subtree plus the 1 root node plus recursively the number of nodes in the right subtree.
• If no, then the last node on the last tree row is in the left subtree and the right subtree is a full tree of height h-2. So we take the 2^(h-1)-1 nodes of the right subtree plus the 1 root node plus recursively the number of nodes in the left subtree.

Since I halve the tree in every recursive step, I have O(log(n)) steps. Finding a height costs O(log(n)). So overall O(log(n)^2).

class Solution {
    int height(TreeNode root) {
        return root == null ? -1 : 1 + height(root.left);
    }
    public int countNodes(TreeNode root) {
        int h = height(root);
        System.out.println(h);
        return h < 0 ? 0 :
               height(root.right) == h-1 ? (1 << h) + countNodes(root.right)
                                         : (1 << h-1) + countNodes(root.left);
    }
}

height-----根的height为0，height是最大子串长度-1

2. brute force（preorder traverse）

class Solution {
    List<Integer> list = new ArrayList();
    public int countNodes(TreeNode root) {
        if(root == null) return list.size();
        list.add(root.val);
        
        if(root.left != null) countNodes(root.left);
        if(root.right != null) countNodes(root.right);
        return list.size();
    }
}

3. 仔细想想完全二叉树，因为特殊性永远至少有一颗子树（或左或右）是满二叉树

利用这个性质，每次我们计算左右子树的高度，如果相当说明是满二叉树，n of nodes = 2 ^ h - 1

如果不相等，我们就返回1 + count(root.left) + count(root.right)，1代表root，这样下去总会有结果

https://leetcode.com/problems/count-complete-tree-nodes/discuss/61948/Accepted-Easy-Understand-Java-Solution

class Solution {
    public int countNodes(TreeNode root) {
        int lh = lefth(root);
        int rh = righth(root);
        
        if(lh == rh) return (1<<lh) - 1;
        else{
            return 1 + countNodes(root.left) + countNodes(root.right);
        }
    }
    public int lefth(TreeNode root){
        int h = 0;
        while(root != null){
            root = root.left;
            h++;
        }
        return h;
    }
    public int righth(TreeNode root){
        int h = 0;
        while(root != null){
            root = root.right;
            h++;
        }
        return h;
    }
}