LeetCode 124:Binary Tree Maximum Path Sum

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;}