leetcode222-完全二叉树的节点个数

222. 完全二叉树的节点个数

这道题如果要最快，就要充分利用完全二叉树的性质。甚至还有二分查找法，还没怎么认真看

利用树的深度判断是否为完全二叉树。若是，直接公式得出节点数。

class Solution {
public:
    int countNodes(TreeNode* root) {
        if(root==nullptr) return 0;
        TreeNode* left=root->left;
        TreeNode* right=root->right;
        int leftDepth=0,rightDepth=0;
        while(left)
        {
            left=left->left;leftDepth++;
        }
        while(right)
        {
            right=right->right;rightDepth++;
        }
        if(leftDepth==rightDepth) return (2<<leftDepth)-1;
        else return countNodes(root->left)+countNodes(root->right)+1;
    }
};

递归法

/**
 * 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==nullptr) return 0;
        if(root->left==nullptr&&root->right==nullptr) return 1;
        return countNodes(root->left)+countNodes(root->right)+1;
    }
};

迭代法，我写的这个比递归法慢很多

class Solution {
public:
    int countNodes(TreeNode* root) {
        if(root==nullptr) return 0;
        queue<TreeNode*> qqq;
        qqq.push(root);
        int res=0;
        while(!qqq.empty())
        {
            int size=qqq.size();
            for(int i=0;i<size;i++)
            {
                TreeNode *t=qqq.front();
                qqq.pop();
                res++;
                if(t->left) qqq.push(t->left);
                if(t->right) qqq.push(t->right);
            }
        }
        return res;
    }
};