leetcode || 124、Binary Tree Maximum Path Sum

problem：

Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example:

Given the below binary tree,

1
/ \
2   3

Return

6

.

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题意：在一棵二叉树中寻找一条路径，使其和最大

thinking：

（1）二叉树寻找一条路径比较难做，没有parent指针更难

（2）采用DFS遍历，从根结点开始

code：

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int calLen(TreeNode *root, int &len)
{
if (root == NULL)
{
len = 0;
return 0;
}

if (root->left == NULL && root->right == NULL)
{
len = root->val;
return root->val;
}

int leftPath, rightPath;
int leftLen;
if (root->left)
leftLen = calLen(root->left, leftPath);
else
{
leftLen = INT_MIN;
leftPath = 0;
}

int rightLen;
if (root->right)
rightLen = calLen(root->right, rightPath);
else
{
rightLen = INT_MIN;
rightPath = 0;
}

len = max(max(leftPath, rightPath) + root->val, root->val);
int maxLen = max(root->val, max(leftPath + rightPath + root->val,
max(leftPath + root->val, rightPath + root->val)));

return max(max(leftLen, rightLen), maxLen);
}

int maxPathSum(TreeNode *root) {
int len;
return calLen(root, len);
}
};