描述

222. 完全二叉树的节点个数 - 力扣（Leetcode）

分析

实现

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import java.util.Queue;
import java.util.LinkedList;

//222. 完全二叉树的节点个数
public class CountCompleteTreeNodes222 {
	public static void main(String[] args){
		Integer[] arr = new Integer[]{1,2,3,4,5,6};
		TreeNode node = MyUtils.buildTree(arr);
		int res = new Solution().countNodes(node);
		System.out.println("这颗完全二叉树的节点个数为：" + res);
	}
}

class Solution {

	//直接完全二叉树的玩法
    public int countNodes(TreeNode root) {
    	//终止条件
    	if (root == null) return 0;
    	TreeNode leftNode = root.left, rightNode = root.right;
    	int leftDepth = 0, rightDepth = 0;
    	//leftDepth
    	while(leftNode != null){
    		leftNode = leftNode.left;
    		leftDepth++;
    	}
    	//rightDepth
    	while(rightNode != null){
    		rightNode = rightNode.right;
    		rightDepth++;
    	}
    	if (leftDepth == rightDepth) {
    		return (2 << leftDepth) - 1;
    	}
    	return countNodes(root.left) + countNodes(root.right) + 1;
    }
    //层序队列的迭代实现
    public int countNodes1(TreeNode root) {
    	//终止条件
    	if (root == null) return 0;
    	Queue<TreeNode> que = new LinkedList<>();
    	que.offer(root);
    	int counts = 0, size;
    	while(!que.isEmpty()){
    		size =  que.size();
    		TreeNode temp;
    		for (int i = 0; i < size; i++) {
    			temp = que.poll();
    			counts++;
    			if(temp.left != null) que.offer(temp.left);
    			if (temp.right != null) que.offer(temp.right);
    		}
    	}
    	return counts;
    }
    //递归实现
    public int countNodes2(TreeNode root) {
   		if(root == null)	return 0;
   		return countNodes2(root.left) + countNodes2(root.right) + 1;
    }

}


class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;

	public TreeNode(){};
	public TreeNode(int val){this.val = val;}
	public TreeNode(int val, TreeNode left, TreeNode right){
		this.val = val;
		this.left = left;
		this.right = right;
	}
}


class MyUtils {
	/**
	 *二叉树的构建
	 * int[] -> 二叉树
	 */
	 public static TreeNode buildTree(Integer[] arr) {
        return buildTreeHelper(arr, 0);
    }

    public static TreeNode buildTreeHelper(Integer[] arr, Integer index) {
        if (index >= arr.length || arr[index] == null) {
            return null;
        }

        TreeNode node = new TreeNode(arr[index]);

        node.left = buildTreeHelper(arr, 2 * index + 1);
        node.right = buildTreeHelper(arr, 2 * index + 2);

        return node;
    }
}

总结

实现 时间复杂度 备注
完全二叉树玩法 O(lgn*lgn) 关键就是(满二叉树) 节点数 = (2 << leftDepth) - 1
层序遍历 O(n) 层序遍历的写法
递归 O(n) 两行代码