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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：枚举 + DFS

我们可以枚举二叉树的每个节点，以该节点为根节点，计算其左右子树的最大深度 \(\textit{l}\) 和 \(\textit{r}\)，则该节点的直径为 \(\textit{l} + \textit{r}\)。取所有节点的直径的最大值即为二叉树的直径。

时间复杂度 \(O(n)\)，空间复杂度 \(O(n)\)。其中 \(n\) 为二叉树的节点个数。

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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) {
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int l = dfs(root.left);
        int r = dfs(root.right);
        ans = Math.max(ans, l + r);
        return 1 + Math.max(l, r);
    }
}

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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 diameterOfBinaryTree(TreeNode* root) {
        int ans = 0;
        auto dfs = [&](this auto&& dfs, TreeNode* root) -> int {
            if (!root) {
                return 0;
            }
            int l = dfs(root->left);
            int r = dfs(root->right);
            ans = max(ans, l + r);
            return 1 + max(l, r);
        };
        dfs(root);
        return ans;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func diameterOfBinaryTree(root *TreeNode) (ans int) {
    var dfs func(root *TreeNode) int
    dfs = func(root *TreeNode) int {
        if root == nil {
            return 0
        }
        l, r := dfs(root.Left), dfs(root.Right)
        ans = max(ans, l+r)
        return 1 + max(l, r)
    }
    dfs(root)
    return
}

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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 ans = 0;
    const dfs = (root: TreeNode | null): number => {
        if (!root) {
            return 0;
        }
        const [l, r] = [dfs(root.left), dfs(root.right)];
        ans = Math.max(ans, l + r);
        return 1 + Math.max(l, r);
    };
    dfs(root);
    return ans;
}

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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 {
    pub fn diameter_of_binary_tree(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut ans = 0;
        fn dfs(root: Option<Rc<RefCell<TreeNode>>>, ans: &mut i32) -> i32 {
            match root {
                Some(node) => {
                    let node = node.borrow();
                    let l = dfs(node.left.clone(), ans);
                    let r = dfs(node.right.clone(), ans);

                    *ans = (*ans).max(l + r);

                    1 + l.max(r)
                }
                None => 0,
            }
        }
        dfs(root, &mut ans);
        ans
    }
}

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/**
 * 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 diameterOfBinaryTree = function (root) {
    let ans = 0;
    const dfs = root => {
        if (!root) {
            return 0;
        }
        const [l, r] = [dfs(root.left), dfs(root.right)];
        ans = Math.max(ans, l + r);
        return 1 + Math.max(l, r);
    };
    dfs(root);
    return ans;
};

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    private int ans;

    public int DiameterOfBinaryTree(TreeNode root) {
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int l = dfs(root.left);
        int r = dfs(root.right);
        ans = Math.Max(ans, l + r);
        return 1 + Math.Max(l, r);
    }
}

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int dfs(struct TreeNode* root, int* ans) {
    if (root == NULL) {
        return 0;
    }
    int l = dfs(root->left, ans);
    int r = dfs(root->right, ans);
    if (l + r > *ans) {
        *ans = l + r;
    }
    return 1 + (l > r ? l : r);
}

int diameterOfBinaryTree(struct TreeNode* root) {
    int ans = 0;
    dfs(root, &ans);
    return ans;
}

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