【LeetCode】Binary Tree Maximum Path Sum-优快云博客

参考链接

http://blog.youkuaiyun.com/aay_dps/article/details/8186372

题目描述

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.

题目分析

找出一条路径和最大的值。

定义F（root）是i的孩子到i这段路中的最大值。那么F（root） = MAX（root->val,F(root->left),F(root->right)）

而过当前结点的最大路径各就是root->val + F(root->left)>0?F(root->left):0+ F(root->right)>0?F(root->right):0

遍历所有结点，就可以求出最大路径和了。

总结

在使用INT_MIN时，不要加一个负数，或减一个正数，否则会溢出。

代码示例

/**
 * 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 maxPathSumUtil(TreeNode* root, int& maxLocal)
    {
        if(root == NULL) return INT_MIN;
        int left = maxPathSumUtil(root->left, maxLocal);
        int right = maxPathSumUtil(root->right, maxLocal);
        int local = root->val;
        if(left > 0) local += left;
        if(right > 0) local += right;
        maxLocal = max(local, maxLocal);
        return max(max(left, right), 0)+root->val;
    }
    int maxPathSum(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == NULL) return 0;
        int maxLocal = root->val;
        maxPathSumUtil(root, maxLocal);
        return maxLocal;
    }
};

class Solution {
public:
    int maxPathSum(TreeNode *root) {
        int maxSum = INT_MIN;
        maxPathSumHelper(root,maxSum);
        return maxSum;
    }
    int maxPathSumHelper(TreeNode *root, int &maxSum)
    {
        if(root == NULL)    return INT_MIN;////////////////////后面不能再加负数
        int left = maxPathSumHelper(root->left,maxSum);
        int right = maxPathSumHelper(root->right,maxSum);
        int tmpmax = root->val;
        if(left>0) tmpmax+=left;
        if(right>0) tmpmax+=right;
        
        maxSum = max(maxSum,tmpmax);
        
        //return max(root->val,max(root->val+left,root->val+right));
        return root->val+max(max(left,right),0);
    }
};