543 Diameter of Binary Tree 二叉树的直径
Description:
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.
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。
思路:
对每个结点求左子树和右子树深度, 取左右子树深度和的最大值
时间复杂度O(n), 空间复杂度O(n)
代码:
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)
{
if (!root) return 0;
int result = 0;
unordered_map<TreeNode*, int> depths;
stack<TreeNode*> s;
s.push(root);
while (s.size())
{
TreeNode* cur = s.top();
if (cur -> left and depths.find(cur -> left) == depths.end()) s.push(cur -> left);
else if (cur -> right and depths.find(cur -> right) == depths.end()) s.push(cur -> right);
else
{
s.pop();
int l = depths[cur -> left], r = depths[cur -> right];
depths[cur] = max(l, r) + 1;
result = max(result, l + r);
}
}
return result;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private int result = 0;
public int diameterOfBinaryTree(TreeNode root) {
depth(root);
return result;
}
private int depth(TreeNode root) {
if (root == null) return 0;
int l = depth(root.left);
int r = depth(root.right);
result = Math.max(l + r, result);
return Math.max(l, r) + 1;
}
}
Python:
# 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: TreeNode) -> int:
if not root:
return 0
result = 0
def depth(root: TreeNode) -> int:
nonlocal result
if not root:
return 0
l, r = depth(root.left), depth(root.right)
result = max(l + r, result)
return max(l, r) + 1
depth(root)
return result
