222. Count Complete Tree Nodes

222. 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.

[思路]

用暴力法, recursive求会超时 O(N).   如果从某节点一直向左的高度 = 一直向右的高度, 那么以该节点为root的子树一定是complete binary tree. 而 complete binary tree的节点数,可以用公式算出 2^h - 1. 如果高度不相等, 则递归调用 return countNode(left) + countNode(right) + 1.  复杂度为O(h^2) 

完全递归的写法：int countNodes(TreeNode* root) {        if(!root)            return 0;        else            return 1+countNodes(root->left)+countNodes(root->right);    }超时，改进后如下：int countNodes(TreeNode* root) {        if(!root) return 0;        int hl=0, hr=0;        TreeNode *l=root, *r=root;        while(l) {hl++;l=l->left;}        while(r) {hr++;r=r->right;}        if(hl==hr) return pow(2,hl)-1;        return 1+countNodes(root->left)+countNodes(root->right);    }