222. Count Complete Tree Nodes

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

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Solution #1, 递归。对比左右子树的 深度，如果相等，则缺口开始在右树，可以把左树的数量全加上。右树类似。
O(lg N * lg N)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        if (root == null) return 0;
        int lh = height(root.left), rh = height(root.right);
        if (lh == rh) {
            return 1 + (1 << lh) - 1 + countNodes(root.right);
        }else {
            return 1 + (1 << rh) - 1 + countNodes(root.left);
        }
    }
    
    private int height(TreeNode root) {
        int h = 0;
        while (root != null) {
            h++;
            root = root.left;
        }
        return h;
    }
}

Solution #2, 迭代写法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        if (root == null) return 0;
        
        int h = height(root) - 1;
        int count = 0;
            
        while (root != null) {
            if (height(root.right) == h - 1) {
                count += 1 + (1 << (h - 1)) - 1;
                root = root.left;
            }else {
                count += 1 + (1 << h) - 1;
                root = root.right;
            }
            h--;
        }
        
        return count;
    }
    
    private int height(TreeNode root) {
        int h = 0;
        while (root != null) {
            h++;
            root = root.left;
        }
        return h;
    }
}