LeetCode 543. Diameter of Binary Tree(二叉树的直径)
题目描述
Given a binary tree, you need to compute 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.
Example:
Given a binary tree
1
/ \
2 3
/ \
4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
Note: The length of path between two nodes is represented by the number of edges between them.
解法一
思路
二叉树上的任一“路径”上一定有一个结点是所有其他结点的祖先结点(因为“路径”是由一个个父子关系连接而成的),那么换个表述方法,对于任一结点,以此结点为根的diameter就可以表示为左子树高度 + 右子树高度 + 1
,而二叉树的diameter就是所有结点为根的diameter中最大的那个。
那么这题实际也是一个二叉树遍历的问题,即对每个结点,计算左子树高度 + 右子树高度 + 1
。那么应该用前序遍历还是后序遍历呢?如果我们把这题再抽象,其实就是一个求二叉树高度的问题,那么显然就是后序遍历了。
Python
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def diameterOfBinaryTree(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.result = -1
def diameter(root):
if root is None:
return 0
left = diameter(root.left)
right = diameter(root.right)
length = left + right + 1
self.result = max(self.result, length)
return max(left, right) + 1
diameter(root)
return 0 if self.result == -1 else self.result - 1
Java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int diameterOfBinaryTree(TreeNode root) {
class Utils {
int result = -1;
int diameter(TreeNode root) {
if (root == null) {
return 0;
}
int left = diameter(root.left);
int right = diameter(root.right);
int length = left + right + 1;
result = Math.max(result, length);
return Math.max(left, right) + 1;
}
}
Utils utils = new Utils();
utils.diameter(root);
return utils.result == -1 ? 0 : utils.result - 1;
}
}
C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int diameterOfBinaryTree(TreeNode *root) {
class Utils {
public:
int result = -1;
int diameter(TreeNode *root) {
if (root == nullptr) {
return 0;
}
int left = diameter(root->left);
int right = diameter(root->right);
int length = left + right + 1;
result = max(result, length);
return max(left, right) + 1;
}
};
Utils utils;
utils.diameter(root);
return utils.result == -1 ? 0 : utils.result - 1;
}
};