題目

https://leetcode.com/problems/longest-univalue-path/

題解：樹形 dp 套路

實現1：帶有 Info 類的樹形 dp

/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/ class Info {int depth;int val; }class Solution {int result;public int longestUnivaluePath(TreeNode root) {dfs(root);return result;}public Info dfs(TreeNode node) {if (node == null) return null;Info leftInfo = dfs(node.left);Info rightInfo = dfs(node.right);Info info = new Info();info.depth = 0;info.val = node.val;int length = 0;if (leftInfo != null && leftInfo.val == node.val) {length += leftInfo.depth + 1;info.depth = leftInfo.depth + 1;}if (rightInfo != null && rightInfo.val == node.val) {length += rightInfo.depth + 1;info.depth = Math.max(info.depth, rightInfo.depth + 1);}result = Math.max(result, length);return info;} }

實現2：不帶 Info 類的樹形 dp

后來發現，Info.val 始終是 node.val，所以根本不用存 Info.val 字段，直接用 node.left.val, node.right.val 代替就可以。

然后，Info 類就只剩下一個字段了，一個字段還有啥用？直接用裸的 int 了。

于是，對上面的代碼進行了改造。改造后：

/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/ class Solution {int result;public int longestUnivaluePath(TreeNode root) {dfs(root);return result;}public int dfs(TreeNode node) {if (node == null) return -1;int leftDepth = dfs(node.left);int rightDepth = dfs(node.right);int depth = 0;int length = 0;if (leftDepth != -1 && node.left.val == node.val) {length += leftDepth + 1;depth = leftDepth + 1;}if (rightDepth != -1 && node.right.val == node.val) {length += rightDepth + 1;depth = Math.max(depth, rightDepth + 1);}result = Math.max(result, length);return depth;} }

總結

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