标签:binary 完全二叉树 统计节点数 二叉树 leetcode
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.
思路:分别计算左右子树,然后返回左右字数节点个数加一
递归法:(超时了)
class Solution {
public:
int countNodes(TreeNode* root) {
if(root == NULL) return 0;
int lcount=0,rcount=0;
if(root->left != NULL){
lcount = countNodes(root->left);
}
if(root->right != NULL){
rcount = countNodes(root->right);
}
return lcount+rcount+1;
}
};
方法二:
先计算左右的深度是否相等,相等则为满二叉树,满二叉树的节点个数为深度的平方减一,即depth^2-1;如果不相等,则递归以同样的方式计算左子树和右子树,并返回两者个数之和加一。
Code(c++):
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int countNodes(TreeNode* root) {
if(root == NULL) return 0;
int ldepth = getLeftDepth(root);
int rdepth = getRightDepth(root);
//return the square of leftdepth -1
if(ldepth == rdepth) return (1 << ldepth) - 1;
return countNodes(root->left)+countNodes(root->right)+1;
}
//compute the depth of left tree
int getLeftDepth(TreeNode *root){
int depth = 0;
while(root){
depth++;
root = root->left;
}
return depth;
}
//compute the depth of right tree
int getRightDepth(TreeNode *root){
int depth = 0;
while(root){
depth++;
root = root->right;
}
return depth;
}
};
Leetcode[222]-Count Complete Tree Nodes
标签:binary 完全二叉树 统计节点数 二叉树 leetcode
原文地址:http://blog.csdn.net/dream_angel_z/article/details/46456465