1、没利用完全二叉树性质的递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
Queue<TreeNode> q = new LinkedList<>();
public int countNodes(TreeNode root) {
if(root == null) return 0;
return countNodes(root.left) + countNodes(root.right) + 1;
}
}
2、因为完全二叉树只有最后一层不是满的。
1.1、左子树不是满二叉树,右子树自然就是满二叉树了
1.2、左子树是满二叉树,右子树不一定。
/**
* 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 left = countLevel(root.left);
int right = countLevel(root.right);
if(left == right){//左子树是满二叉树
return countNodes(root.right) + (1<<left);//左子树加上根节点数目刚好是2^left,用位运算快一点
}else{
return countNodes(root.left) + (1<<right);//同理
}
}
private int countLevel(TreeNode root){//可以帮助判断是否左子树是满二叉树
int level = 0;
while(root != null){
level++;
root = root.left;
}
return level;
}
}