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543. 二叉树的直径

English Version

题目描述

给定一棵二叉树，你需要计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过也可能不穿过根结点。

 

示例 :
给定二叉树

          1
         / \
        2   3
       / \     
      4   5    

返回 3, 它的长度是路径 [4,2,1,3] 或者 [5,2,1,3]。

 

注意：两结点之间的路径长度是以它们之间边的数目表示。

解法

类似题目：687. 最长同值路径

Python3

# 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:
        res = 0

        def dfs(root):
            nonlocal res
            if root is None:
                return 0
            left, right = dfs(root.left), dfs(root.right)
            res = max(res, left + right)
            return 1 + max(left, right)
        
        dfs(root)
        return res

Java

/**
 * 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 res;

    public int diameterOfBinaryTree(TreeNode root) {
        res = 0;
        dfs(root);
        return res;
    }

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

C++

/**
 * 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 res;

    int diameterOfBinaryTree(TreeNode* root) {
        res = 0;
        dfs(root);
        return res;
    }

    int dfs(TreeNode* root) {
        if (!root) return 0;
        int left = dfs(root->left), right = dfs(root->right);
        res = max(res, left + right);
        return 1 + max(left, right);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
var res int

func diameterOfBinaryTree(root *TreeNode) int {
	res = 0
	dfs(root)
	return res
}

func dfs(root *TreeNode) int {
	if root == nil {
		return 0
	}
	left, right := dfs(root.Left), dfs(root.Right)
	res = max(res, left+right)
	return 1 + max(left, right)
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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