[抄题]:
Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root.
Note: The length of path between two nodes is represented by the number of edges between them.
Example 1:
Input:
5
/ \
4 5
/ \ \
1 1 5
Output:
2
Example 2:
Input:
1
/ \
4 5
/ \ \
4 4 5
Output:
2
[暴力解法]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
- 因为可以不从root开始,以为要分情况:进行后续计算之后发现可以合并:从头开始肯定比较长,没必要讨论
- 以为左右两边也要分开讨论:结果求和加一下就行了,还是见得太少
[一句话思路]:
点在线段的dfs上加一,没地方直接加,需要单独写公式
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
Example:
...
/
4 (res = resl + resr = 3)
(resl = 2) / \ (resr= 1)
(l = 1) 4 4 (r = 0)
/
4
点数是线段数+1
[一刷]:
- DFS是嵌套traverse的过程,不能当作返回值,加深理解一下
- DFS求的是左边或右边单独的最大值,不是合集,稍微理解下 res[0]是所有值中的最大,需要比较,理解题目
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
DFS求的是单边最大值
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
【重磅】java传递的是引用而不是数值,所以必须是数组才有用
[关键模板化代码]:
三元运算符把DFS特殊情况、扩展直接写了,头次见
int resl = (root.left != null && root.val == root.left.val) ? l + 1 : 0;
int resr = (root.right != null && root.val == root.right.val) ? r + 1 : 0;
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int longestUnivaluePath(TreeNode root) {
//corner case
if (root == null) {
return 0;
}
int[] res = new int[1];
//dfs
dfs(root, res);
//return res[0];
return res[0];
} public int dfs(TreeNode root, int[] res) {
//int max = 0;
int l = (root.left != null) ? dfs(root.left, res) : 0;
int r = (root.right != null) ? dfs(root.right, res) : 0;
//if root == root.left, +1
int resl = (root.left != null && root.val == root.left.val) ? l + 1 : 0;
int resr = (root.right != null && root.val == root.right.val) ? r + 1 : 0;
//res[0] is sum
res[0] = Math.max(res[0], resl + resr);
//return the bigger one of l, r
return Math.max(resl, resr);
}
}