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binary-tree-maximum-path-sum

binary-tree-maximum-path-sum

题目描述

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

Return6.

思想

这明显是一道递归题,但是递归不是那么简单,注意返回的值,和最大值的区别

代码

class Solution {
public:
    int maxPathSum(TreeNode *root) {
        if(root == NULL) return 0;
        int max = INT_MIN;
        recurse(root, max);
        return max;
    }

    int recurse(TreeNode *root, int &max) {
        int lMax = 0, rMax = 0;
        int value = root->val;

        if(root->left != NULL) {
           lMax = recurse(root->left, max);
           if(lMax > 0) value += lMax;
        }

        if(root->right != NULL) {
            rMax = recurse(root->right, max);
            if(rMax > 0) value += rMax;
        }

        if(value > max) {
            max = value;
        }
        return Max(root->val, root->val + Max(lMax, rMax));
    }

    int Max(int a, int b) {
        if(b > a) return b;
        else return a;
    }
};
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