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.
最简单的办法是遍历 粗暴遍历会超时
直接遍历每一个节点是不现实的,所以必须通过完全二叉树的特点来计算,我们可以想到,除了最下的那一层,其余的部分,都是满二叉树,这样我们首先可以判断当前的二叉树是不是满二叉树,判断的方法是判断树最左和最右两边的长度是否相等,如果相等就可以直接计算,如果不等就进入递归,分别计算左右两颗子树,知道只有一个节点的时候就停止。
因为完全二叉树进行左右分割之后,很容易就会出现满二叉树,所以节省了大量的遍历节点的时间
因为完全二叉树进行左右分割之后,很容易就会出现满二叉树,所以节省了大量的遍历节点的时间
1 class Solution { 2 public int countNodes(TreeNode root) { 3 if(root==null) return 0; 4 int l = count_l(root); 5 int r = count_r(root); 6 if(l==r) return (1<<l)-1; 7 return 1 + countNodes(root.left) + countNodes(root.right); 8 9 } 10 private int count_l(TreeNode root){ 11 if(root == null) return 0; 12 int cnt = 0; 13 while(root!=null){ 14 cnt+=1; 15 root =root.left; 16 } 17 return cnt; 18 } 19 private int count_r(TreeNode root){ 20 if(root == null) return 0; 21 int cnt = 0; 22 while(root!=null){ 23 cnt+=1; 24 root = root.right; 25 } 26 return cnt; 27 } 28 }