[LeetCode] 222. Count Complete Tree Nodes

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;
    }
};

参考

http://www.cnblogs.com/grandyang/p/4567827.html