[LeetCode] Count Complete Tree Nodes

Count Complete Tree Nodes

Given a complete binary tree, count the number of nodes.

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.

對於完全二叉樹，去掉最后一層，就是一棵滿二叉樹，我們知道高度為 h 的滿二叉樹結點的個數為 2^h - 1 個，所以要知道一棵完全二叉樹的結點個數，只需知道最后一層有多少個結點。而完全二叉樹最后一層結點是從左至右連續的，所以我們可以依次給它們編一個號，然后二分搜索最后一個葉子結點。我是這樣編號的，假設最后一層在 h 層，那么一共有 2^(h-1) 個結點，一共需要 h - 1 位來編號，從根結點出發，向左子樹走編號為 0， 向右子樹走編號為 1，那么最后一層的編號正好從0 ~ 2^(h-1) - 1。復雜度為 O(h*log(2^(h-1))) = O(h^2)。下面是AC代碼。

/**
 * 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:
    bool isOK(TreeNode *root, int h, int v) {
        TreeNode *p = root;
        for (int i = h - 2; i >= 0; --i) {
            if (v & (1 << i)) p = p->right;
            else p = p->left;
        }
        return p != NULL;
    }
    
    int countNodes(TreeNode* root) {
        if (root == NULL) return 0;
        TreeNode *p = root;
        int h = 0;
        while (p != NULL) {
            p = p->left;
            ++h;
        }
        int l = 0, r = (1 << (h - 1)) - 1, m;
        while (l <= r) {
            m = l + ((r - l) >> 1);
            if (isOK(root, h, m)) l = m + 1;
            else r = m - 1;
        }
        return (1 << (h - 1)) + r;
    }
};

 

或者可以用遞歸的方法，對於這個問題，如果從某節點一直向左的高度 = 一直向右的高度, 那么以該節點為root的子樹一定是complete binary tree. 而 complete binary tree的節點數,可以用公式算出 2^h - 1. 如果高度不相等, 則遞歸調用 return countNode(left) + countNode(right) + 1.  復雜度為O(h^2)。

該方法參考：http://blog.csdn.net/xudli/article/details/46385011

/**
 * 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 getHeight(TreeNode *root, bool tag) {
        int h = 0;
        while (root != NULL) {
            if (tag) root = root->left;
            else root = root->right;
            ++h;
        }
        return h;
    }
    
    int countNodes(TreeNode* root) {
        if (root == NULL) return 0;
        int left = getHeight(root, true);
        int right = getHeight(root, false);
        if (left == right) return (1 << left) - 1;
        else return countNodes(root->left) + countNodes(root->right) + 1;
    }
};