README.md

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

方法一：后序遍历求每个结点的深度，此过程中获取每个结点左右子树的最长伸展（深度），迭代获取最长路径。

相似题目：687. 最长同值路径

方法二：构建图，两次 DFS。

相似题目：1245. 树的直径, 1522. N 叉树的直径

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:
        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.
# 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

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 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);
    }
}

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 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);
    }
};

Go

/**
 * 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
}

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

...