题目

给定义一个二叉树, 求二叉树的子路径的最大和.

Input: [1,2,3]
   1
  / \
 2   3
Output: 6
Input: [-10,9,20,null,null,15,7]
  -10
  / \
 9  20
   /  \
  15   7
Output: 42

思路

递归. 分别对左右子树递归.

int maxPathSumHelper(TreeNode* node, int& res) {
    if (node == NULL) return 0;
    int leftMax = max(maxPathSumHelper(node->left, res), 0);
    int rightMax = max(maxPathSumHelper(node->right, res), 0);
    res = max(res, node->val+leftMax+rightMax);
    return node->val + max(leftMax,rightMax);
}

int maxPathSum(TreeNode* root) {
    int res = INT_MIN;
    maxPathSumHelper(root, res);
    return res;
}

总结

递归求最大值, 需要理清思路. 递归程序一看就懂, 但自己写就很难.