给定一个二叉树的 root ，返回 最长的路径的长度 ，这个路径中的 每个节点具有相同值 。 这条路径可以经过也可以不经过根节点。
两个节点之间的路径长度 由它们之间的边数表示。
示例 1:
输入：root = [5,4,5,1,1,5] 输出：2
示例 2:
输入：root = [1,4,5,4,4,5] 输出：2
提示:
- 树的节点数的范围是
[0, 104] -1000 <= Node.val <= 1000- 树的深度将不超过
1000
相似题目：
# 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: TreeNode) -> int:
def dfs(root):
if root is None:
return 0
left, right = dfs(root.left), dfs(root.right)
left = left + 1 if root.left and root.left.val == root.val else 0
right = right + 1 if root.right and root.right.val == root.val else 0
nonlocal ans
ans = max(ans, left + right)
return max(left, right)
ans = 0
dfs(root)
return ans/**
* 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 {
private int ans;
public int longestUnivaluePath(TreeNode root) {
ans = 0;
dfs(root);
return ans;
}
private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int left = dfs(root.left);
int right = dfs(root.right);
left = root.left != null && root.left.val == root.val ? left + 1 : 0;
right = root.right != null && root.right.val == root.val ? right + 1 : 0;
ans = Math.max(ans, left + right);
return Math.max(left, right);
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int ans;
int longestUnivaluePath(TreeNode* root) {
ans = 0;
dfs(root);
return ans;
}
int dfs(TreeNode* root) {
if (!root) return 0;
int left = dfs(root->left), right = dfs(root->right);
left = root->left && root->left->val == root->val ? left + 1 : 0;
right = root->right && root->right->val == root->val ? right + 1 : 0;
ans = max(ans, left + right);
return max(left, right);
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func longestUnivaluePath(root *TreeNode) int {
ans := 0
var dfs func(root *TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
left, right := dfs(root.Left), dfs(root.Right)
if root.Left != nil && root.Left.Val == root.Val {
left++
} else {
left = 0
}
if root.Right != nil && root.Right.Val == root.Val {
right++
} else {
right = 0
}
ans = max(ans, left+right)
return max(left, right)
}
dfs(root)
return ans
}/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function longestUnivaluePath(root: TreeNode | null): number {
if (root == null) {
return 0;
}
let res = 0;
const dfs = (root: TreeNode | null, target: number) => {
if (root == null) {
return 0;
}
const { val, left, right } = root;
let l = dfs(left, val);
let r = dfs(right, val);
res = Math.max(res, l + r);
if (val === target) {
return Math.max(l, r) + 1;
}
return 0;
};
dfs(root, root.val);
return res;
}// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, target: i32, res: &mut i32) -> i32 {
if root.is_none() {
return 0;
}
let root = root.as_ref().unwrap().as_ref().borrow();
let left = Self::dfs(&root.left, root.val, res);
let right = Self::dfs(&root.right, root.val, res);
*res = (*res).max(left + right);
if root.val == target {
return left.max(right) + 1;
}
0
}
pub fn longest_univalue_path(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
if root.is_none() {
return 0;
}
let mut res = 0;
Self::dfs(&root, root.as_ref().unwrap().as_ref().borrow().val, &mut res);
res
}
}/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var longestUnivaluePath = function (root) {
let ans = 0;
let dfs = function (root) {
if (!root) {
return 0;
}
let left = dfs(root.left),
right = dfs(root.right);
left = root.left?.val == root.val ? left + 1 : 0;
right = root.right?.val == root.val ? right + 1 : 0;
ans = Math.max(ans, left + right);
return Math.max(left, right);
};
dfs(root);
return ans;
};/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
#define max(a, b) (((a) > (b)) ? (a) : (b))
int dfs(struct TreeNode* root, int target, int* res) {
if (!root) {
return 0;
}
int left = dfs(root->left, root->val, res);
int right = dfs(root->right, root->val, res);
*res = max(*res, left + right);
if (root->val == target) {
return max(left, right) + 1;
}
return 0;
}
int longestUnivaluePath(struct TreeNode* root) {
if (!root) {
return 0;
}
int res = 0;
dfs(root, root->val, &res);
return res;
}