https://oj.leetcode.com/problems/binary-tree-maximum-path-sum/ 求二叉树的连续路径加和的最大值。 这道题会被坑的地方就是认为结点值都大于0。需要考虑结点值小于0的情况。 考虑包含结点u的所有路径的最大值。可以看出这个最大路径有三种可能: 1)只含u自己一个结点。root->val 2)包含u和一个儿子,但...
}//对每个节点遍历求左右两个节点的做大加上本身,然后取最大的值就是maximum path sum了intmaxPathSum(TreeNode *root) {if(!root)return0;inttmpl = INT_MIN, tmpr =INT_MIN;intcur =curMax(root);if(root ->left) tmpl= maxPathSum(root ->left);if(root ->right) tmpr= maxPathSum(root ->...
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example: Given the below binary tree andsum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1 1. 2. 3. 4. 5. 6. ...
要满足从 root 到叶节点的和为 sum, 只需要满足从左/右孩子到叶节点的和为 sum - root->val 即可。 AC 代码 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left;[](h * TreeNode *right; * Tre...
Left, remainSum) { return true } // 当右子结点存在,且存在一条右子结点到叶子路径上所有值到和为 remainSum ,则满足题意 if root.Right != nil && hasPathSum(root.Right, remainSum) { return true } // 左右子结点都不满足题意,返回 false return false } 题目链接: Path Sum: leetcode.com...
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...
http://bangbingsyb.blogspot.com/2014/11/leetcode-binary-tree-maximum-path-sum.html http://www.programcreek.com/2013/02/leetcode-binary-tree-maximum-path-sum-java/ 另外,这道题目的BST条件,似乎没什么用。因为如果全是负数,BST也没帮助了。
publicintMaxPathSum(TreeNode root) { MaxPath(root); return_max; } privateintMaxPath(TreeNode current) { if(current == null) return0; // 如果子树路径和为负则应当置0表示最大路径不包含子树 intleft = Math.Max(MaxPath(current.left),0); ...
# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = NoneclassSolution:def__init__(self):self.result=[]defpathSum(self,root:TreeNode,sum:int)->bool:# DFS & recursion |defrecursion(root,path,cur_sum):if...
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