Leetcode balancetree

最新推荐文章于 2022-05-26 10:45:05 发布

weixin_30724853

最新推荐文章于 2022-05-26 10:45:05 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/tgkx1054/archive/2013/04/16/3025324.html

本文提供了两个关于二叉树的经典算法实现：一是判断二叉树是否平衡的方法；二是寻找二叉树中最大路径和的解决方案。通过递归方式深入探讨了二叉树的性质，并展示了具体的C++代码实现。

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class Solution {
public:
    bool isBalanced(TreeNode *root) {
        if(root==NULL)return true;
        int left=depth(root->left);
        int right=depth(root->right);
        if(abs(left-right)>1)return false;
        return isBalanced(root->left)&&isBalanced(root->right);       
    }
    int depth(TreeNode *root)
    {
        if(root==NULL)return 0;
        int left=depth(root->left);
        int right=depth(root->right);
        return left>right?left+1:right+1;
    }
};

Max PathSum

class Solution {
public:
    vector<int>v;
     
    int maxPathSum(TreeNode *root) {     
        v.clear();
        PreOrder(root);
        int max=v[0];
        for(int i=0;i<v.size();i++)
        {
            if(max<v[i])max=v[i];
        }
        return max;             
    }
    void PreOrder(TreeNode *root)
    {
        v.push_back(TwomaxSum(root));
        if(root->left)PreOrder(root->left);
        if(root->right)PreOrder(root->right);        
    }
    int TwomaxSum(TreeNode *root)
    {
        if(root==NULL)return 0;
        int sum=root->val;
        if(maxSum(root->left)>0)sum+=maxSum(root->left);
        if(maxSum(root->right)>0)sum+=maxSum(root->right);
        return sum;
    }
    int maxSum(TreeNode *root)
    {
        if(root==NULL)return 0;
        if(root->left==NULL&&root->right==NULL)return root->val;
        int left=maxSum(root->left);
        int right=maxSum(root->right);
        if(left<=0&&right<=0)return root->val;
        else return left>=right?(left+root->val) :(right+root->val);
    }
};

转载于:https://www.cnblogs.com/tgkx1054/archive/2013/04/16/3025324.html