[LeetCode222]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.

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Hide Similar Problems (E) Closest Binary Search Tree Value

這道題費了不少時間。
首先complete binary tree 就是除了最後一層可能不滿，其他層都是滿的。 其次如果一個binary tree heigh 是h，則它的node一共有pow（2，h）－1。這個冪指數函數又等價於 (1<<h)。知道這幾點，我們就知道總的node 數爲

(1<<(height−1))−1+cnt

where cnt is the nodes of last level.

下面這個code最快，還需要好好理解一下，基本就是tmp找到了最後一個node的位置。。

/**
 * 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) return 0;
        TreeNode *temp = root;
        int height = 0, count = 0, level;
        while(temp) {
            temp = temp->left;
            height ++;
        }
        temp = root;
        level = height - 2;
        while(level >= 0) {
            TreeNode *left = temp->left;
            for(int i = 0;i < level;i ++) {
                left = left->right;
            }
            if(left) {//which means there might be more node in right sub part, then tmp goes rightward.
                temp = temp->right;
                count += (1 << level);
            } else temp = temp->left;
            level --;
        }
        if(temp) count ++;
        return (1 << (height - 1)) + count - 1;
    }
};

傳統recursion比較好理解， 但時間複雜度較高: O(h^2).

/**
 * 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) return 0;
        TreeNode* l = root;
        TreeNode* r = root;
        int cntL = 0, cntR = 0;
        while(l){
            ++cntL;
            l = l->left;
        }
        while(r){
            ++cntR;
            r = r->right;
        }
        if(cntL == cntR) return (1 << cntL) - 1;
        else return 1 + countNodes(root->left) + countNodes(root->right);
    }
};