Binary Tree Maximum Path Sum (LeetCode)

Question:

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

 

解答：

这问题求的是任何一个节点它的左边最大Path和右边最大Path的和的最大值。对于任意一个node来说，它的最大Path Sum无非是以下几种情况中的一种：

1. node左子树中的最大Path Sum

2. node右子树中的最大Path Sum

3. 左子树到node的最大Path(max(max path of node->left + node, node))和右子树到node的最大Path(max(max path of node->right + node, node))

所以递归函数不仅要返回最大Path Sum，也要返回当前这个node所能形成的最大Path。

注意左右子树不存在的情况，以及Path Sum可能是负数的情况。

 

    int maxPathSum(TreeNode *root) {
        
        if (!root)
            return 0;
            
        int maxPath = 0;
        int maxSum = 0;
        
        FindMaxPathSum(root, maxPath, maxSum);
        
        return maxSum;
    }
    
    void FindMaxPathSum(TreeNode* root, int& maxPath, int& maxSum)
    {
        maxSum = root->val;
        
        int leftMaxPath = 0;
        int rightMaxPath = 0;
        int childMaxSum = 0;
        
        if (root->left)
        {
            FindMaxPathSum(root->left, leftMaxPath, childMaxSum);
            maxSum = max(childMaxSum, maxSum);
        }
        
        if (root->right)
        {
            FindMaxPathSum(root->right, rightMaxPath, childMaxSum);
            maxSum = max(maxSum, childMaxSum);
        }
        
        maxPath = root->val;
        
        if (max(leftMaxPath, rightMaxPath) > 0)
            maxPath += max(leftMaxPath, rightMaxPath);
        
        int thisSum = root->val + (leftMaxPath > 0 ? leftMaxPath : 0) + (rightMaxPath > 0 ? rightMaxPath : 0);
        
        maxSum = max(maxSum, thisSum);
    }

 

之前以为是求任何一棵子树的所有点的和的最大值，不过方法一样 (未验证)。

    // 最大子树的和，可能包含负数，子树的叶节点可能是任意节点（不一定是原来的叶节点），如果都是非负数，则是所有节点的和。
    void FindMaxSubTreeSum(TreeNode* root, int& maxSumWithRoot, int& maxSum)
    {
        maxSumWithRoot = root->val;
        maxSum = root->val;
        
        int childMaxSumWithRoot = 0;
        int childMaxSum = 0;
        
        if (root->left)
        {
            FindMaxPathSum(root->left, childMaxSumWithRoot, childMaxSum);
            
            maxSumWithRoot = max(maxSumWithRoot, childMaxSumWithRoot + maxSumWithRoot);
            maxSum = max(childMaxSum, maxSumWithRoot);
        }
        
        if (root->right)
        {
            FindMaxPathSum(root->right, childMaxSumWithRoot, childMaxSum);
            
            maxSumWithRoot = max(maxSumWithRoot, childMaxSumWithRoot + maxSumWithRoot);
            maxSum = max(childMaxSum, maxSumWithRoot);
        }
    }