给你一棵二叉树的根节点，返回该树的 直径 。
二叉树的 直径 是指树中任意两个节点之间最长路径的 长度 。这条路径可能经过也可能不经过根节点 root 。
两节点之间路径的 长度 由它们之间边数表示。
示例 1：
输入：root = [1,2,3,4,5] 输出：3 解释：3 ，取路径 [4,2,1,3] 或 [5,2,1,3] 的长度。
示例 2：
输入：root = [1,2] 输出：1
提示：
- 树中节点数目在范围
[1, 104]内 -100 <= Node.val <= 100
方法一：后序遍历求每个结点的深度，此过程中获取每个结点左右子树的最长伸展（深度），迭代获取最长路径。
相似题目：687. 最长同值路径
方法二：构建图，两次 DFS。
相似题目：1245. 树的直径, 1522. N 叉树的直径
# 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 diameterOfBinaryTree(self, root: TreeNode) -> int:
def dfs(root):
if root is None:
return 0
nonlocal ans
left, right = dfs(root.left), dfs(root.right)
ans = max(ans, left + right)
return 1 + max(left, right)
ans = 0
dfs(root)
return ans# 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 diameterOfBinaryTree(self, root: TreeNode) -> int:
def build(root):
if root is None:
return
nonlocal d
if root.left:
d[root].add(root.left)
d[root.left].add(root)
if root.right:
d[root].add(root.right)
d[root.right].add(root)
build(root.left)
build(root.right)
def dfs(u, t):
nonlocal ans, vis, d, next
if u in vis:
return
vis.add(u)
if t > ans:
ans = t
next = u
for v in d[u]:
dfs(v, t + 1)
d = defaultdict(set)
ans = 0
next = root
build(root)
vis = set()
dfs(next, 0)
vis.clear()
dfs(next, 0)
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 diameterOfBinaryTree(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);
ans = Math.max(ans, left + right);
return 1 + 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 diameterOfBinaryTree(TreeNode* root) {
ans = 0;
dfs(root);
return ans;
}
int dfs(TreeNode* root) {
if (!root) return 0;
int left = dfs(root->left);
int right = dfs(root->right);
ans = max(ans, left + right);
return 1 + max(left, right);
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func diameterOfBinaryTree(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)
ans = max(ans, left+right)
return 1 + max(left, right)
}
dfs(root)
return ans
}
func max(a, b int) int {
if a > b {
return a
}
return b
}/**
* 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* res) {
if (!root) {
return 0;
}
int left = dfs(root->left, res);
int right = dfs(root->right, res);
*res = max(*res, left + right);
return max(left, right) + 1;
}
int diameterOfBinaryTree(struct TreeNode* root) {
int res = 0;
dfs(root, &res);
return res;
}/**
* 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 diameterOfBinaryTree(root: TreeNode | null): number {
let res = 0;
const dfs = (root: TreeNode | null) => {
if (root == null) {
return 0;
}
const { left, right } = root;
const l = dfs(left);
const r = dfs(right);
res = Math.max(res, l + r);
return Math.max(l, r) + 1;
};
dfs(root);
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>>>, 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, res);
let right = Self::dfs(&root.right, res);
*res = (*res).max(left + right);
left.max(right) + 1
}
pub fn diameter_of_binary_tree(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let mut res = 0;
Self::dfs(&root, &mut res);
res
}
}