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543. 二叉树的直径

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

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# 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

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/**
 * 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);
    }
}

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/**
 * 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);
    }
};

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/**
 * 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
}

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/**
 * 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;
}

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// 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::cell::RefCell;
use std::rc::Rc;
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
    }
}

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/**
 * 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

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# 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

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