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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,
1 / 2 3
Return 6
.
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//找出二叉树任意一点到另一点的路径,使得和最大. //解题思路:后序遍历,先计算左右子树的值l和r,若l<0或r<0,则不用加上 class Solution { public: int maxSum; int maxPathSum(TreeNode* root) { if (root == NULL) return 0; maxSum =root->val; dfs(root); return maxSum; } int dfs(TreeNode* root){ if (root == NULL) return 0; int l = dfs(root->left); int r = dfs(root->right); int sum = root->val; if (l > 0) sum = sum + l; if (r > 0) sum = sum + r; maxSum = max(maxSum, sum); return max(root->val, max(l + root->val, r + root->val)); } };
#include<iostream> #include<new> #include<vector> #include<algorithm> using namespace std; //Definition for binary tree struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; //找出二叉树任意一点到另一点的路径,使得和最大. //解题思路:后序遍历,先计算左右子树的值l和r,若l<0或r<0,则不用加上 class Solution { public: int maxSum; int maxPathSum(TreeNode* root) { if (root == NULL) return 0; maxSum =root->val; dfs(root); return maxSum; } int dfs(TreeNode* root){ if (root == NULL) return 0; int l = dfs(root->left); int r = dfs(root->right); int sum = root->val; if (l > 0) sum = sum + l; if (r > 0) sum = sum + r; maxSum = max(maxSum, sum); return max(root->val, max(l + root->val, r + root->val)); } }; void createTree(TreeNode *&root) { int i; cin >> i; if (i != 0) { root = new TreeNode(i); if (root == NULL) return; createTree(root->left); createTree(root->right); } } int main() { Solution s; TreeNode *root; createTree(root); int sum = s.maxPathSum(root); cout << sum << endl; system("pause"); return 0; }
LeetCode 124:Binary Tree Maximum Path Sum
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原文地址:http://blog.csdn.net/geekmanong/article/details/50437766