}intmaxPathSum(TreeNode *root) {if(root == NULL)return0;intcsum =INT_MIN;intmaxsum =INT_MIN; maxpathsumhelper(root, csum, maxsum);returnmaxsum; }
接着我们发现上述计算max 和 求出MAX的过程完全可以放到func(node) 里去。 按照这个思路的代码,maxPathSumCore 就是上面 func(node)的实现: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), r...
Given a binary tree, find the maximum path sum. For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root. For example: Given the below binary tree, ...
int maxPathSum(TreeNode *root) { max_sum = INT_MIN; dfs(root); return max_sum; } int dfs1(const TreeNode *root) { if (root == nullptr)return 0; int l = dfs1(root->left); int r = dfs1(root->right); int sum = root->val; if (l > 0)sum += l; if (r > 0)sum ...
public int maxPathSum(TreeNode root) { if (root == null) { return Integer.MIN_VALUE; } //左子树的最大值 int left = maxPathSum(root.left); //右子树的最大值 int right = maxPathSum(root.right); //再考虑包含根节点的最大值 int all = ...; return Math.max(Math.max(left, right...
Given a non-empty binary tree, find the maximum path sum. For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root...
# Definition for a binary tree node.# class TreeNode(object):# def __init__(self, x):# self.val = x# self.left = None# self.right = NoneclassSolution(object):defmaxPathSum(self,root):""":typeroot:TreeNode:rtype:int"""self.maxSum=0self.findMax(root)returnself.maxSumdeffindMax...
- 1996 () Citation Context ...efficient parallel program for the tree version of the maximum segment sum problem [20]. Several studies have been done on the parallelization of the problem: on lists [3,21], on 2-dimensional arrays =-=[22]-=-, and on binary trees [18]. To the best...
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For example, the probability of observing total copy number 3 is a sum over all compatible haplotype-specific copy numbers (0, 3), (1, 2), (2, 1), and (3, 0). In other words, the haplotype-specific copy numbers are latent variables, and the likelihood is an average over them. ...