Leetcode[124] Binary Tree Maximum Path Sum
Given a 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
The path must contain at least one node and does not need
go through the root.
For example: Given the below binary tree,
1 / \ 2 3
Return 6.
DFS
复杂度
O(N)
思路
对于每一节点,考虑到这一个节点为止,所能形成的最大值。Math.max(left.val, right.val) + root.val,是经过这个节点为止的能形成的最大值的一条支路。
代码
int sum = Integer.MIN_VALUE;
public int maxPathSum(TreeNode root) {
if(root == null) return 0;
if(root.left == null && root.right == null) return root.val;
int left = Math.max(0, maxPathSum(root.left));
int right = Math.max(0, maxPathSum(root.right));
sum = Math.max(sum, left + right + root.val);
return Math.max(left, right) + root.val;
}
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