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.

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完全二叉树。思路是先找到树的高度h，如果右子树的高度为h-1的话证明最后一个节点在右子树，左子树为满树，因此res+=左子树的节点个数+根节点=2^(h-1)，根节点移到右边。

如果右子树的高度不为h-1的话证明最后一个节点在左子树，右子树为满树，res+=右子树的节点个数+根节点=2^(h-2).根节点移到左边。

class Solution {public:    int height(TreeNode* root){        return root==NULL ? -1 : height(root->left)+1;    }    int countNodes(TreeNode* root) {                int sum=0, h=height(root);        while(root){            if(height(root->right) == h - 1){                sum += pow(2,h);                root = root->right;            }            else{                sum += pow(2,h-1);                root = root->left;            }            h--;        }        return sum;    }};

需要注意的是这里为了计算方便，直接返回的h值就是实际的h值-1.比如叶子节点的高度是0.