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543. Diameter of Binary Tree

Description

Given the root of a binary tree, return the length of the diameter of the tree.

The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

The length of a path between two nodes is represented by the number of edges between them.

 

Example 1:

Input: root = [1,2,3,4,5]
Output: 3
Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].

Example 2:

Input: root = [1,2]
Output: 1

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • -100 <= Node.val <= 100

Solutions

Solution 1: Enumeration + DFS

Thinking

The diameter is the longest number of edges between any two nodes. Pairwise paths are unrealistic. Any diameter passes through some node as the sum of that node's two subtree heights.

DFS returns height and, on the way up, updates a global answer with \(l+r\), then returns \(1+\max(l,r)\) to the parent. Each node is visited once.

We can enumerate each node of the binary tree, and for each node, calculate the maximum depth of its left and right subtrees, \(\textit{l}\) and \(\textit{r}\), respectively. The diameter of the node is \(\textit{l} + \textit{r}\). The maximum diameter among all nodes is the diameter of the binary tree.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the number of nodes in the binary tree.

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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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