Question

543. 二叉树的直径

English Version

题目描述

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

解法

方法一

# 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.
 * 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
}

/**
 * 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
  }
}

/**
 * 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;
}

方法二

# 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