代码随想录：222、完全二叉树的结点个数

222、完全二叉树的结点个数

层序遍历法

比较简单，但时间复杂度高

/**
 * 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) {
        int sum=0;
        Queue<TreeNode> q=new LinkedList<>();
        if(root==null)return sum;
        q.add(root);
        while(!q.isEmpty())
        {
            TreeNode cur=q.poll();
            sum++;
            if(cur.left!=null)q.add(cur.left);
            if(cur.right!=null)q.add(cur.right);
        }
        return sum;
    }
}

递归

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

递归+树高度

时间复杂度O(Log² N)
如果一边为满二叉树，能用2的树高度次幂算出树的结点个数+1个父结点。

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