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
Approach #1: C++.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int maxPathSum(TreeNode* root) {
if (!root) return 0;
int ans = INT_MIN;
solve(root, ans);
return ans;
}
private:
int solve(TreeNode* root, int& ans) {
if (!root) return 0;
int l = max(0, solve(root->left, ans));
int r = max(0, solve(root->right, ans));
int sum = l + r + root->val;
ans = max(ans, sum);
return max(l, r) + root->val;
}
};
Approach #2: Java.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
int ans = Integer.MIN_VALUE;
public int maxPathSum(TreeNode root) {
if (root == null) return 0;
solve(root);
return ans;
}
private int solve(TreeNode root) {
if (root == null) return 0;
int l = Math.max(0, solve(root.left));
int r = Math.max(0, solve(root.right));
int sum = l + r + root.val;
ans = Math.max(ans, sum);
return Math.max(l, r) + root.val;
}
}
Approach #3: Python.
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def solve(self, root):
if not root: return 0
l = max(0, self.solve(root.left))
r = max(0, self.solve(root.right))
self.ans = max(self.ans, l + r + root.val)
return max(l, r) + root.val
def maxPathSum(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root == None:
return 0
self.ans = -sys.maxint
self.solve(root)
return self.ans