222. Count Complete Tree Nodes

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原文链接：https://segmentfault.com/a/1190000018226297

文章标签：

#java #数据结构与算法 #runtime

本文介绍了一种使用后序遍历算法来统计完全二叉树中节点数量的方法。通过递归地计算左子树和右子树的节点数，并加上根节点，实现了高效的节点计数。该算法在Java实现中表现优异，运行速度快于100%的在线提交。

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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

难度：medium

题目：给定完全二叉树，统计其结点数。

思路：后序遍历

Runtime: 1 ms, faster than 100.00% of Java online submissions for Count Complete Tree Nodes.
Memory Usage: 40.4 MB, less than 45.43% of Java online submissions for Count Complete Tree Nodes.

/**
 * 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 (null == root) {
            return 0;
        }
        
        return countNodes(root.left) + countNodes(root.right) + 1;
    }
}