Given a complete binary tree, count the number of nodes.
Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.
解法一：递归超时
代码如下：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int countNodes(TreeNode root) {
if(root==null)return 0;
int num1=0,num2=0;
num1=count(root.left);
num2=count(root.right);
return 1+num1+num2; }
int count(TreeNode root){
if(root==null) return 0;
int num1=0,num2=0;
if(root.left==null && root.right==null)
return 1;
if(root.left!=null)
num1=count(root.left);
if(root.right!=null)
num2=count(root.right);
return 1+num1+num2;
}
}
运行结果：
解法二：层次遍历，队列,超时
代码如下：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int countNodes(TreeNode root) {
if(root==null)return 0; Queue<TreeNode> q=new LinkedList<>();
q.offer(root);
int num=1;
TreeNode tmp=null;
while(!q.isEmpty()){
tmp=q.poll();
if(tmp.left!=null){
q.offer(tmp.left);
num++;
}
if(tmp.right!=null){
q.offer(tmp.right);
num++;
}
}
return num;
} }
解法三：如果从某节点一直向左的高度 = 一直向右的高度, 那么以该节点为root的子树一定是complete binary tree. 而 complete binary tree的节点数,可以用公式算出 2^h - 1. 如果高度不相等, 则递归调用 return countNode(left) + countNode(right) + 1. 复杂度为O(h^2)
代码如下：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int countNodes(TreeNode root) {
if(root==null) return 0; int l = getLeft(root) + 1;
int r = getRight(root) + 1; if(l==r) {
return (2<<(l-1)) - 1;
} else {
return countNodes(root.left) + countNodes(root.right) + 1;
}
} private int getLeft(TreeNode root) {
int count = 0;
while(root.left!=null) {
root = root.left;
++count;
}
return count;
} private int getRight(TreeNode root) {
int count = 0;
while(root.right!=null) {
root = root.right;
++count;
}
return count;
}
}
运行结果：
