[leedcode 222] Count Complete Tree Nodes
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.
/** * 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) { //适合任何一颗二叉树 /*方法一: if(root==null) return 0; return countNodes(root.left)+countNodes(root.right)+1;*/ /* 方法二: if(root==null) return 0; LinkedList<TreeNode> queue=new LinkedList<TreeNode>(); queue.add(root); int count=0; while(!queue.isEmpty()){ int temp=queue.size(); count+=temp; for(int i=0;i<temp;i++){ TreeNode node=queue.remove(); if(node.left!=null){ queue.add(node.left); count++; } if(node.right!=null){ queue.add(node.right); count++; } } } return count;*/ /*可以证明一个完全二叉树左右子树至少有一个是满二叉树。 满二叉树的节点数是2^k-1,k是树的深度。 所以我们可以先判断该树是否为满二叉树,然后是的话直接返回结果,如果不是递归地求解子树。 这样不用遍历所有的节点。复杂度小于O(N),比对所有点遍历复杂度要小,最好的情况是O(lgN)。 推算大概在O(lgN)~O(N)之间。*/ if(root==null) return 0; TreeNode l=root; TreeNode r=root; int leftH=0; int rightH=0; while(l!=null){ l=l.left; leftH++; } while(r!=null){ r=r.right; rightH++; } if(leftH==rightH){ return (1<<leftH)-1;//注意优先级 }else{ return countNodes(root.left)+countNodes(root.right)+1; } } }