1、沒利用完全二叉樹性質的遞歸
/**
* 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、因為完全二叉樹只有最后一層不是滿的。
1.1、左子樹不是滿二叉樹,右子樹自然就是滿二叉樹了
1.2、左子樹是滿二叉樹,右子樹不一定。
/**
* 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){//左子樹是滿二叉樹
return countNodes(root.right) + (1<<left);//左子樹加上根節點數目剛好是2^left,用位運算快一點
}else{
return countNodes(root.left) + (1<<right);//同理
}
}
private int countLevel(TreeNode root){//可以幫助判斷是否左子樹是滿二叉樹
int level = 0;
while(root != null){
level++;
root = root.left;
}
return level;
}
}