TreeNode *pNodeA2 = CreateBinaryTreeNode(6); TreeNode *pNodeA3 = CreateBinaryTreeNode(1); TreeNode *pNodeA4 = CreateBinaryTreeNode(9); TreeNode *pNodeA5 = CreateBinaryTreeNode(2); TreeNode *pNodeA6 = CreateBinar
}privateResultType helper(TreeNode root) {if(root ==null) {returnnewResultType(Integer.MIN_VALUE, Integer.MIN_VALUE); //两种路径中都最少包括一个结点 }//DivideResultType left =helper(root.left); ResultType right=helper(root.right);//Conquerintroot2Any =Math.max(0, Math.max(left.root2Any,...
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...
* public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { private int max = Integer.MIN_VALUE; public int maxPathSum(TreeNode root) { helper(root); return max; } public int helper(TreeNode root) { if...
分析:http:///2013/01/leetcode-binary-tree-maximum-path-sum.html For each node like following, there should be four ways existing for max path: 1. Node only (因为本题中的节点可能是负值!) 2. L-sub + Node 3. R-sub + Node
1. Maximum Depth of Binary Tree 解法1: class Solution: def maxDepth(self, root: TreeNode) -> int: self.answer = 0 self.helper(root,0) return self.answer def helper(self,node,depth): if node is None: self.answer=max(self.answer,depth) else: self.helper(node.left,depth+1) self....
* TreeNode(int x) { val = x; } * } */publicclassSolution{publicintmaxPathSum(TreeNoderoot){int[]max=newint[1];max[0]=Integer.MIN_VALUE;findMax(root,max);returnmax[0];}privateintfindMax(TreeNoderoot,int[]max){if(root==null)return0;intleft=findMax(root.left,max);intright=find...
这道题比较简单,因为题目提示了是求从根节点出发的最大路径和,只需要记录一条当前path, 随时更新一个maxSum, 用递归和回溯就可以解决了。 publicclassSolution{/* * @param root: the root of binary tree. * @return: An integer */intmaxSum=Integer.MIN_VALUE;publicintmaxPathSum2(TreeNoderoot){// ...
connections. The path must contain at least one node and does not need to go through the root. 举例: Given the below binary tree, 1 / \ 2 3 Return 6. 题目分析: 找从任意节点出发的任意路径的最大长度。 每个node都有可能是其他路径上的node,这种情况要max(left,right)。如此循环。 每个node都...
def maxPathSum(self, root): ''' :type root: TreeNode :rtype: int ''' self.maxes = -float('inf')def helper(root): myValue = root.valif not root.left and not root.right: self.maxes = myValue if myValue > self.maxes else self.maxes ...