[leetcode] 124. Binary Tree Maximum Path Sum

Given a binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root.

For example:

Given the below binary tree,

       1
      / \     2   3

Return6.

这道题是计算二叉树最大路径和，题目难度为Hard。

基于二叉树数据结构的特性，我们知道最大路径和的路径肯定存在一个自己的根节点，它的左右子树（形似链表）即是路径的左右两部分，这点需要大家首先确认。这样我们以每个节点作为此根节点，分别计算从它的左右子树开始的最大路径和，之后加上此根节点值即是此根节点确定的最大路径和。这里需要注意如果左右子树的最大路径和是负数，则抛弃这个子树的路径，因为加上它会使路径和变小，把这个子树的路径和记为0即可。这样深度优先遍历二叉树即可比较获得最大路径和。具体代码：

class Solution {    int getMaxPathSum(TreeNode* root, int& maxSum) {        if(!root) return 0;        int left = max(getMaxPathSum(root->left, maxSum), 0);        int right = max(getMaxPathSum(root->right, maxSum), 0);        maxSum = max(maxSum, left+right+root->val);        return max(left, right) + root->val;    }public:    int maxPathSum(TreeNode* root) {        if(!root) return 0;        int maxSum = INT_MIN;        getMaxPathSum(root, maxSum);        return maxSum;    }};