LeetCode: Binary Tree Maximum Path Sum 解题报告
Binary Tree Maximum Path Sum
Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example:
Given the below binary tree,
1
/ \
2 3
SOLUTION 1:
计算树的最长path有2种情况:
1. 通过根的path.
(1)如果左子树从左树根到任何一个Node的path大于零,可以链到root上
(2)如果右子树从右树根到任何一个Node的path大于零,可以链到root上
2. 不通过根的path. 这个可以取左子树及右子树的path的最大值。
所以创建一个inner class:
记录2个值:
1. 本树的最大path。
2. 本树从根节点出发到任何一个节点的最大path.
注意,当root == null,以上2个值都要置为Integer_MIN_VALUE; 因为没有节点可取的时候,是不存在solution的。以免干扰递归的计算
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public class ReturnType {
int maxSingle;
int max;
ReturnType (int maxSingle, int max) {
this.max = max;
this.maxSingle = maxSingle;
}
} public int maxPathSum(TreeNode root) {
return dfs(root).max;
} public ReturnType dfs(TreeNode root) {
ReturnType ret = new ReturnType(Integer.MIN_VALUE, Integer.MIN_VALUE);
if (root == null) {
return ret;
} ReturnType left = dfs(root.left);
ReturnType right = dfs(root.right); int cross = root.val; // if any of the path of left and right is below 0, don't add it.
cross += Math.max(0, left.maxSingle);
cross += Math.max(0, right.maxSingle); // 注意,这里不可以把Math.max(left.maxSingle, right.maxSingle) 与root.val加起来,
// 会有可能越界!
int maxSingle = Math.max(left.maxSingle, right.maxSingle); // may left.maxSingle and right.maxSingle are below 0
maxSingle = Math.max(maxSingle, 0);
maxSingle += root.val; ret.maxSingle = maxSingle;
ret.max = Math.max(right.max, left.max);
ret.max = Math.max(ret.max, cross); return ret;
}
}
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/MaxPathSum.java
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