解题报告 - LeetCode 222. 完全二叉树的节点

LeetCode 222. 完全二叉树的节点个数

@TOC

题目描述

 给你一棵 完全二叉树 的根节点 root ，求出该树的节点个数。

完全二叉树 的定义如下：在完全二叉树中，除了最底层节点可能没填满外，其余每层节点数都达到最大值，并且最下面一层的节点都集中在该层最左边的若干位置。若最底层为第 h 层，则该层包含 1~ 2h 个节点。

示例：[图片上传失败...(image-ca28f6-1665618553274)]

输入：root = [1,2,3,4,5,6]
输出：6

提示：

树中节点的数目范围是[0, 5 * 104]
0 <= Node.val <= 5 * 104
题目数据保证输入的树是 完全二叉树

一、解题关键词

完全二叉树
节点个数

二、解题报告

1.思路分析

1. 树的第一反映 要递归（1、隐藏内部细节 2、重复性计算过程）
2. 左子树 + 右子树 + 根节点

2.时间复杂度

3.代码示例

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        //树 使用递归
        if(null == root) return 0;
        int left = countNodes(root.left);
        int right = countNodes(root.right);

        return left + right +1;

    }
}

代码二：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        if(null == root){
            return 0;
        }
        int left = countLevel(root.left);
        int right = countLevel(root.right);

        if(left == right){
            return countNodes(root.right) +(1<< left);
        }else{
            return countNodes(root.left) + (1<<right);
        }
    }
    int countLevel(TreeNode root){
        int level = 0;
        while(root != null){
            level++;
            root = root.left;
        }
        return level;
    }
}

代码三：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        if(null == root){
            return 0;
        }
        TreeNode node = root;
        int leftH = 0;
        int rightH = 0;
        while(null != node){
            leftH ++;
            node = node.left;
        }
        while(null != node){
            rightH ++;
            node = node.right;
        }
        if(leftH == rightH){
            return (int) Math.pow(2,leftH + 1) -1;
        }
        return countNodes(root.left) + countNodes(root.right) + 1;
    }
}

4.知识点

1<< left == (int) Math.pow(2,leftH + 1) -1