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
给一个非空二叉树，找出最大路径和。
解法：递归，与常规path sum不同，这题的path sum可以不起始于root，也可以不终止于leaf。
Java:
public class Solution {
int maxValue;
public int maxPathSum(TreeNode root) {
maxValue = Integer.MIN_VALUE;
maxPathDown(root);
return maxValue;
}
private int maxPathDown(TreeNode node) {
if (node == null) return 0;
int left = Math.max(0, maxPathDown(node.left));
int right = Math.max(0, maxPathDown(node.right));
maxValue = Math.max(maxValue, left + right + node.val);
return Math.max(left, right) + node.val;
}
}
Python:
# Time: O(n)
# Space: O(h), h is height of binary tree
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution(object):
maxSum = float("-inf")
# @param root, a tree node
# @return an integer
def maxPathSum(self, root):
self.maxPathSumRecu(root)
return self.maxSum
def maxPathSumRecu(self, root):
if root is None:
return 0
left = max(0, self.maxPathSumRecu(root.left))
right = max(0, self.maxPathSumRecu(root.right))
self.maxSum = max(self.maxSum, root.val + left + right)
return root.val + max(left, right)
C++:
class Solution {
public:
int maxPathSum(TreeNode* root) {
int res = INT_MIN;
helper(root, res);
return res;
}
int helper(TreeNode* node, int& res) {
if (!node) return 0;
int left = max(helper(node->left, res), 0);
int right = max(helper(node->right, res), 0);
res = max(res, left + right + node->val);
return max(left, right) + node->val;
}
};
Follow up: 返回最大路径，递归函数就不能返回路径和，而是返回该路径上所有结点组成的数组，递归的参数还要保留最大路径和，对左右子节点调用递归函数后得到的是数组，统计出数组之和，并且跟0比较，如果小于0，和清零，数组清空。
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