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

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;
    }
};