python3 解法, 执行用时: 348 ms, 内存消耗: 19.1 MB, 提交时间: 2022-09-02 10:56:19

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def longestUnivaluePath(self, root: Optional[TreeNode]) -> int:
        ans = 0
        def dfs(node: Optional[TreeNode]) -> int:
            if node is None:
                return 0
            left = dfs(node.left)
            right = dfs(node.right)
            left1 = left + 1 if node.left and node.left.val == node.val else 0
            right1 = right + 1 if node.right and node.right.val == node.val else 0
            nonlocal ans
            ans = max(ans, left1 + right1)
            return max(left1, right1)
        dfs(root)
        return ans

golang 解法, 执行用时: 52 ms, 内存消耗: 7.3 MB, 提交时间: 2022-09-02 10:55:35

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func longestUnivaluePath(root *TreeNode) (ans int) {
    var dfs func(*TreeNode) int
    dfs = func(node *TreeNode) int {
        if node == nil {
            return 0
        }
        left := dfs(node.Left)
        right := dfs(node.Right)
        left1, right1 := 0, 0
        if node.Left != nil && node.Left.Val == node.Val {
            left1 = left + 1
        }
        if node.Right != nil && node.Right.Val == node.Val {
            right1 = right + 1
        }
        ans = max(ans, left1+right1)
        return max(left1, right1)
    }
    dfs(root)
    return ans
}

func max(a, b int) int {
    if b > a {
        return b
    }
    return a
}