The path may startandend at any node in the tree. For example: Given the below binary tree, 1 / \ 2 3 Return 6. 2.解法分析: leetcode中给出的函数头为:int maxPathSum(TreeNode *root) 给定的数据结构为: Definitionforbinary tree *structTreeNode { *intval; * TreeNode *left; * TreeNo...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public:intmaxPathSum(TreeNode *root) { maxPathSumCore(root);returnMAX; }intmaxPathSumCore(TreeNode *node) {if(NULL == node)return0;inta = maxPathSumCore(node ->left);intb = maxPathSumCore(node -...
取其4个值中得最大值作为子树的最大路径和。 classSolution {public:intmaxSum =INT_MIN;intgetMaxSum(TreeNode*root){if(root == NULL)return0;intleftSum = getMaxSum(root->left), rightSum = getMaxSum(root->right);intcurSum = max(root->val, max(root->val+leftSum, root->val+rightSum...
max);returnmax[0];}publicintcalculateSum(TreeNoderoot,int[]max){if(root==null)return0;intleft=calculateSum(root.left,max);intright=calculateSum(root.right,max);intcurrent=Math.max(root.val,Math.max(root.val+left,root
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。
答案 classSolution{int globalMax=Integer.MIN_VALUE;publicintmaxPathSum(TreeNoderoot){recur(root);returnglobalMax;}publicintrecur(TreeNoderoot){if(root==null)return0;// Cut negative partintleft=Math.max(0,recur(root.left));intright=Math.max(0,recur(root.right));int localMax=left+right+roo...
Binary Tree Maximum Path Sum@LeetCode Binary Tree Maximum Path Sum 动态规划+深度优先搜索。把大问题(求整棵树的路径最大值)拆分成小问题(每颗子树的路径最大值),递推公式为:当前树的路径最大值=max(左子树的路径最大值, 右子树的路径最大值)+当前根节点的值。以此来推出最后全树的最大路径值。
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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...
参考: https://shenjie1993.gitbooks.io/leetcode-python/124%20Binary%20Tree%20Maximum%20Path%20Sum.html 我们现在要求最大路径和,那么就要分别得到左右两条路径的最大和。而左路径的最大和为左节点的值加上它左右路径中较大的路径和,右路径最大和为右节点的值加上它左右路径中较大的路径和。 注意:如果...