LeetCode-124. Binary Tree Maximum Path Sum

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

题解：

后序遍历搜索左右子树最大序列，每次更新节点的单向（左or右子树）最长路径，然后保存最大值。

class Solution {
public:
  void getMaxPathSum(TreeNode *root, int &ans) {
      if (root == NULL) {
          return;
      }
      getMaxPathSum(root->left, ans);
      getMaxPathSum(root->right, ans);
      int leftVal = 0, rightVal = 0, fullVal = 0;
      if (root->right == NULL && root->left == NULL) {
        ans= max(ans, root->val);
      }
      else if (root->right != NULL && root->left != NULL) {
        leftVal = root->val + root->left->val;
        rightVal = root->val + root->right->val;
        fullVal = root->val + root->right->val + root->left->val;
        ans = max({ans, fullVal, leftVal, rightVal});
      }
      else if (root->left != NULL && root->right == NULL) {
        leftVal = root->val + root->left->val;
        ans = max(ans, leftVal);
      }
      else if (root->right != NULL && root->left == NULL) {
        rightVal = root->val + root->right->val;
        ans = max(ans, rightVal);
      }
      root->val = max({root->val, leftVal, rightVal});
  }
  int maxPathSum(TreeNode* root) {
    if (root == NULL) {
      return 0;
    }
    int ans = root->val;
    getMaxPathSum(root, ans);
    return ans;
  }
};