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.

 

 

class Solution {
public:
    int countNodes(TreeNode* root) {
    
        if (root == NULL) return 0;
        int lh = 0;
        int rh = 0;
        auto l = root;
        auto r = root;
        while(l!=NULL) {lh++;l=l->left;}
        while(r!=NULL) {rh++;r=r->right;}
        if (lh==rh) return pow(2,lh)-1;
        return 1 + countNodes(root->left) + countNodes(root->right);
    return 0;
    }
};

 

 

 

 

 

 

最简单的办法是遍历 粗暴遍历会超时

直接遍历每一个节点是不现实的，所以必须通过完全二叉树的特点来计算，我们可以想到，除了最下的那一层，其余的部分，都是满二叉树，这样我们首先可以判断当前的二叉树是不是满二叉树，判断的方法是判断树最左和最右两边的长度是否相等，如果相等就可以直接计算，如果不等就进入递归，分别计算左右两颗子树，知道只有一个节点的时候就停止。
因为完全二叉树进行左右分割之后，很容易就会出现满二叉树，所以节省了大量的遍历节点的时间

 

 

 

 

class Solution {
    public int countNodes(TreeNode root) {
        if(root==null) return 0;
        int l = count_l(root);
        int r = count_r(root);
        if(l==r) return (1<<l)-1;
        return 1 + countNodes(root.left) + countNodes(root.right);
        
    }
    private int count_l(TreeNode root){
        if(root == null) return 0;
        int cnt = 0;
        while(root!=null){
            cnt+=1;
            root =root.left;
        }
        return cnt;
    }
    private int count_r(TreeNode root){
        if(root == null) return 0;
        int cnt = 0;
        while(root!=null){
            cnt+=1;
            root = root.right;
        }
        return cnt;
    }
}