222. Count Complete Tree Nodes (Easy)

Given the root of a complete binary tree, return the number of the nodes in the tree.

According to Wikipedia, every level, except possibly the last, is completely filled in a complete binary tree, 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.

Design an algorithm that runs in less than O(n) time complexity.

 

Example 1:

Input: root = [1,2,3,4,5,6]
Output: 6

Example 2:

Input: root = []
Output: 0

Example 3:

Input: root = [1]
Output: 1

 

Constraints:

• The number of nodes in the tree is in the range [0, 5 * 104].
• 0 <= Node.val <= 5 * 104
• The tree is guaranteed to be complete.

A:

----------------------Solution 2 ----------------------
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)  

 public class Solution {  
    public int countNodes(TreeNode root) {  
        if(root==null) return 0;  
          
        int l = getLeft(root) + 1;  
        int r = getRight(root) + 1;  
          
        if(l==r) {  
            return (2<<(l-1)) - 1;  
        } else {  
            return countNodes(root.left) + countNodes(root.right) + 1;  
        }  
    }  
      
    private int getLeft(TreeNode root) {  
        int count = 0;  
        while(root.left!=null) {  
            root = root.left;  
            ++count;  
        }  
        return count;  
    }  
      
    private int getRight(TreeNode root) {  
        int count = 0;  
        while(root.right!=null) {  
            root = root.right;  
            ++count;  
        }  
        return count;  
    }  
}  

***********上面的是每次查询了最左边和最右边的高度. 还不够好************

********     3rd pass,   每次对比左右子树的高度   如果相等， 则左边的为full bi nary tree **** 否则，右边的也是full binary tree 但是少了一层*****

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int countNodes(TreeNode* root) {
        if(!root)
            return 0;
        int hLeft = getHeight(root->left);
        int hRight = getHeight(root->right);
        if( hRight == hLeft){ //right child height == left child height 
            return (1<< hLeft) + countNodes(root->right);
        }else{
            return (1 << hLeft-1) + countNodes(root->left);
        }
    }
private:
    int getHeight(TreeNode* root){ // get height for complete tree, just goes down to the left most node
        if(!root)
            return 0;
        return 1 + getHeight(root->left);
    }
};

上面的需要计算2次高度， 也是浪费。 因为左边的只需要计算一次就可以了

class Solution {

public:

int countNodes(TreeNode* root) {

return countNodesWithDepth(root, getHeight(root));

}

private:

int countNodesWithDepth(TreeNode* root, int height){

if(!root)

return 0;

if( getHeight(root->right) == height-1){ //right child height == left child height

return (1 << height-1) + countNodesWithDepth(root->right, height-1);

}else{

return (1 << height-2) + countNodesWithDepth(root->left, height-1);

}

}

int getHeight(TreeNode* root){ // get height for complete tree, just goes down to the left most node

if(!root)

return 0;

return 1 + getHeight(root->left);

}

};

但是，理论上是少算了一次，但是LC上并没有速度的改进，反而内存稍微增加了点儿。

Mistakes：
需要用 移位运算， 如果用Math.pow(a,b)的话，会超时