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

求完整二叉樹的節點個數，直接遞迴左右節點個數。分別找出以當前節點為根節點的左子樹和右子樹的高度並對比，如果相等，則說明是滿二叉樹，直接返回節點個數，如果不相等，則節點個數為左子樹的節點個數加上右子樹的節點個數再加1(根節點)，其中左右子樹節點個數的計算可以使用遞迴來計算。

程式碼

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;
    }
};

參考

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