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

  1. The path must contain at least one node and does not need

  2. 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;
}

hellolittleJ17
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