LeetCode—***寻找二叉树中任意两个节点之间的最大值Binary Tree Maximum Path Sum

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Binary Tree Maximum Path Sum

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.

这是一个二叉树的题目，首先应该了解二叉树树结构，寻找最大路径，从left,right节点递归处理

这里用一个全局变量，记录left+root+right的最大值

之前对递归算法很不熟悉，其实只要将一层递归逻辑理清楚，然后有退出的条件，就可以了

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int max = -9999;      int FindMax(TreeNode * root)    {        int value = root->val;        int leftValue = 0;        int rightValue = 0;        int rt;        if(root == NULL)        {            return 0;        }        if(root->left != NULL)        {            leftValue = FindMax(root->left);            if(leftValue > 0)            value += leftValue;        }        if(root->right != NULL)        {            rightValue = FindMax(root->right);            if(rightValue > 0)            value += rightValue;        }        if(value > max)        {            max = value;           //<用max记录root+left+right的最大值        }        if(root->val > (root->val+leftValue))        {            rt = root->val;        }        else        {            rt = root->val+leftValue;        }        if(rt < (root->val+rightValue))   //<返回root,root+left.root+right中的最大值        {            rt = root->val+rightValue;        }        return rt;            }    int maxPathSum(TreeNode *root)     {        if(NULL == root)        {            return 0;        }        FindMax(root);        return max;    }};

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