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
思路
dfs:
2
/
1 -3
maxPath: 1 + 2 + 0
1. use maxPath to update result from all paths which is max path sum
2. divide a path into 3 parts:
a. left path: from root to left side down to 0 or more steps
b. right path: from root to right side down to 0 or more steps
c. cur root itself
3. maxPath = c + (a>0 ? a: 0 ) + (b > 0 ? b:0)
这是一道关于BST和recursion的经典题,需要掌握
最naive的想法是找到所有BST的path,返回max
发现, 任意一条path都有一个顶点(位置最高点)
我们用这个顶点来分解所有path
这样,以任意一个点为顶点的path就分解为
a. max_sum (左边path)
b. max_sum (右边path)
c. 顶点自己的value
进一步,
a + b + c 组成的人字形path的max path sum
2 / 1 -3
dfs的return value : 2(顶点自己的value必须加上,无论正负) + 1 (正数贡献自己) + 0 (-3为负数不做贡献就是及时止损了) = 3
跟 [leetcode]543. Diameter of Binary Tree二叉树直径 的思路基本一致。
代码
1 class Solution { 2 public int maxPathSum(TreeNode root) { 3 // corner case 4 if(root == null){return 0;} 5 /* 要么用个global variable放在class下,要么用长度为1的一维数组来存。 6 maxSum的value,可正可负,初始化为Integer.MIN_VALUE。 7 */ 8 int[] maxPath = new int[]{Integer.MIN_VALUE}; 9 dfs(root, maxPath); 10 return maxPath[0]; 11 } 12 // 递归求以root为顶点所有直上直下的path中,path sum最大的一条值。没有U-turn的 13 private int dfs(TreeNode root, int[]maxPath){ 14 // left > 0 benefit sum, add to sum; left < 0 will worsen sum, default 0 15 int left = root.left != null ? Math.max(dfs(root.left, maxPath), 0) : 0; 16 int right = root.right != null ? Math.max(dfs(root.right, maxPath), 0) : 0; 17 int cur = root.val + left + right; 18 maxPath[0] = Math.max(maxPath[0], cur); 19 return root.val + Math.max(left, right); 20 } 21 }