每日一道算法题--leetcode 222--彻底二叉树的节点个数--python&C++

时间 2019-11-07

标签每日一道算法 leetcode 彻底二叉树节点个数 python&c python 栏目应用数学繁體版

原文原文链接

【题目描述】python

【思路】

1.一种是递归 2.非递归则使用层序遍历，队列实现bash

【python代码】app

1.递归
class Solution:
    def countNodes(self, root: TreeNode) -> int:
        if root:
            return 1+self.countNodes(root.left)+self.countNodes(root.right)
        else:
            return 0
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2.非递归
class Solution:
    def countNodes(self, root: TreeNode) -> int:
        l=[]
        re=0
        if root:
            l.append(root)
        while l:
            first=l.pop(0)
            re+=1
            if first.left:
                l.append(first.left)
            if first.right:
                l.append(first.right)
        return re
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【C++代码】ui

1.递归：
class Solution {
public:
    int countNodes(TreeNode* root) {
        return root==NULL?0:1+countNodes(root->left)+countNodes(root->right);
    }
};
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2.非递归
class Solution {
public:
    int countNodes(TreeNode* root) {
        if(root==NULL)
            return 0;
        queue <TreeNode*> q;
        int re=0;
        q.push(root);
        while(!q.empty()){
            TreeNode* t=q.front();
            q.pop();
            re+=1;
            if(t->left!=NULL){q.push(t->left);}
            if(t->right!=NULL){q.push(t->right);}
        }
        return re;       
        }
};
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