Count Complete Tree Nodes
Given a complete binary tree, count the number of nodes.
Note:
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.
Example:
Input:
1
/ \
2 3
/ \ /
4 5 6
Output: 6
解析
求完整二叉树的节点个数,直接递归左右节点个数。分别找出以当前节点为根节点的左子树和右子树的高度并对比,如果相等,则说明是满二叉树,直接返回节点个数,如果不相等,则节点个数为左子树的节点个数加上右子树的节点个数再加1(根节点),其中左右子树节点个数的计算可以使用递归来计算。
代码
class Solution {
public:
int countNodes(TreeNode* root) {
if(!root) return 0;
TreeNode* leftnode = root->left;
TreeNode* rightnode = root->right;
int lefth,righth;
lefth = righth = 0;
while(leftnode){
lefth ++;
leftnode = leftnode->left;
}
while(rightnode){
righth ++;
rightnode = rightnode->right;
}
if(lefth == righth) return pow(2, lefth+1) -1;
else return countNodes(root->left) + countNodes(root->right) +1;
}
};