C++实现LeetCode(124.求二叉树的最大路径和)

[LeetCode] 124. Binary Tree Maximum Path Sum 求二叉树的最大路径和

Given a non-empty 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 must contain at least one node and does not need to go through the root.

Example 1:

Input: [1,2,3]

       1
/ \
2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
/ \
9  20
/  \
15   7

Output: 42

这道求二叉树的最大路径和是一道蛮有难度的题，难就难在起始位置和结束位置可以为任意位置，博主当然是又不会了，于是上网看看大神们的解法，像这种类似树的遍历的题，一般来说都需要用 DFS 来求解，先来看一个简单的例子：

    4
/ \   11 13
/ \ 7   2

由于这是一个很简单的例子，很容易就能找到最长路径为 7->11->4->13，那么怎么用递归来找出正确的路径和呢？根据以往的经验，树的递归解法一般都是递归到叶节点，然后开始边处理边回溯到根节点。这里就假设此时已经递归到结点7了，其没有左右子节点，如果以结点7为根结点的子树最大路径和就是7。然后回溯到结点 11，如果以结点 11 为根结点的子树，最大路径和为 7+11+2=20。但是当回溯到结点4的时候，对于结点 11 来说，就不能同时取两条路径了，只能取左路径，或者是右路径，所以当根结点是4的时候，那么结点 11 只能取其左子结点7，因为7大于2。所以，对于每个结点来说，要知道经过其左子结点的 path 之和大还是经过右子节点的 path 之和大。递归函数返回值就可以定义为以当前结点为根结点，到叶节点的最大路径之和，然后全局路径最大值放在参数中，用结果 res 来表示。

在递归函数中，如果当前结点不存在，直接返回0。否则就分别对其左右子节点调用递归函数，由于路径和有可能为负数，这里当然不希望加上负的路径和，所以和0相比，取较大的那个，就是要么不加，加就要加正数。然后来更新全局最大值结果 res，就是以左子结点为终点的最大 path 之和加上以右子结点为终点的最大 path 之和，还要加上当前结点值，这样就组成了一个条完整的路径。而返回值是取 left 和 right 中的较大值加上当前结点值，因为返回值的定义是以当前结点为终点的 path 之和，所以只能取 left 和 right 中较大的那个值，而不是两个都要，参见代码如下：

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

讨论：这道题有一个很好的 Follow up，就是返回这个最大路径，那么就复杂很多，因为这样递归函数就不能返回路径和了，而是返回该路径上所有的结点组成的数组，递归的参数还要保留最大路径之和，同时还需要最大路径结点的数组，然后对左右子节点调用递归函数后得到的是数组，要统计出数组之和，并且跟0比较，如果小于0，和清零，数组清空。然后就是更新最大路径之和跟数组啦，还要拼出来返回值数组，代码长了很多，有兴趣的童鞋可以在评论区贴上你的代码～

Github 同步地址：

https://github.com/grandyang/leetcode/issues/124

类似题目：

Path Sum

Sum Root to Leaf Numbers

Path Sum IV

Longest Univalue Path

参考资料：

https://leetcode.com/problems/binary-tree-maximum-path-sum/

https://leetcode.com/problems/binary-tree-maximum-path-sum/discuss/39775/Accepted-short-solution-in-Java

https://leetcode.com/problems/binary-tree-maximum-path-sum/discuss/39869/Simple-O(n)-algorithm-with-one-traversal-through-the-tree

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