Binary Tree Maximum Path Sum(leetcode)

题目：

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1      / \     2   3

Return 6.

题目来源：https://oj.leetcode.com/problems/binary-tree-maximum-path-sum/

解题思路：用后序遍历进行，想访问左右孩子节点，然后访问根节点，不断更新根节点的val值，不断向上遍历，直到root节点。此时保存的maxResult值就是所求。但是这种方法改变了原来节点的值，如果不能改变原来节点值，则可以考虑用深搜的办法《leetcode题解中给出了答案》

#include<iostream>using namespace std;struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};void postOrderTraverse(TreeNode *root,int &maxResult){int leftChildValue=0,rightChildValue=0;if(root->left!=NULL){postOrderTraverse(root->left,maxResult);leftChildValue=root->left->val;}if(root->right!=NULL){postOrderTraverse(root->right,maxResult);rightChildValue=root->right->val;}int sum=root->val;if(leftChildValue>0)sum+=leftChildValue;if(rightChildValue>0)sum+=rightChildValue;maxResult=max(maxResult,sum);root->val+=max(leftChildValue,rightChildValue)>0?max(leftChildValue,rightChildValue):0;}int maxPathSum(TreeNode *root){if(root==NULL)return 0;int result=INT_MIN;postOrderTraverse(root,result);return result;}int main(){TreeNode *root=new TreeNode(2);root->left=new TreeNode(-1);//root->right=new TreeNode(3);cout<<maxPathSum(root)<<endl;system("pause");return 0;}

《leetcode题解》中用深搜来解这个题，其实这个方法在hihocoder的每周测试中已经用过了。

#include<iostream>using namespace std;struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};int dfs(TreeNode *root,int &maxResult){if(root==NULL)return 0;int l=dfs(root->left,maxResult);int r=dfs(root->right,maxResult);int sum=root->val;if(l>0)sum+=l;if(r>0)sum+=r;maxResult=max(maxResult,sum);return max(r,l)>0?max(r,l)+root->val:root->val;}int maxPathSum(TreeNode *root){if(root==NULL)return 0;int result=INT_MIN;dfs(root,result);return result;}int main(){TreeNode *root=new TreeNode(2);root->left=new TreeNode(-1);//root->right=new TreeNode(3);cout<<maxPathSum(root)<<endl;system("pause");return 0;}