Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root.
Note: The length of path between two nodes is represented by the number of edges between them.
Example 1:
Input:
5
/ \
4 5
/ \ \
1 1 5
Output:
2
Example 2:
Input:
1
/ \
4 5
/ \ \
4 4 5
Output:
2
Note: The given binary tree has not more than 10000 nodes. The height of the tree is not more than 1000.
给一个二叉树，找出最长的相同值路径，与250. Count Univalue Subtrees类似。
解法：递归
Java:
class Solution {
public int longestUnivaluePath(TreeNode root) {
int[] res = new int[1];
if (root != null) dfs(root, res);
return res[0];
}
private int dfs(TreeNode node, int[] res) {
int l = node.left != null ? dfs(node.left, res) : 0; // Longest-Univalue-Path-Start-At - left child
int r = node.right != null ? dfs(node.right, res) : 0; // Longest-Univalue-Path-Start-At - right child
int resl = node.left != null && node.left.val == node.val ? l + 1 : 0; // Longest-Univalue-Path-Start-At - node, and go left
int resr = node.right != null && node.right.val == node.val ? r + 1 : 0; // Longest-Univalue-Path-Start-At - node, and go right
res[0] = Math.max(res[0], resl + resr); // Longest-Univalue-Path-Across - node
return Math.max(resl, resr);
}
}
Python:
# Time: O(n)
# Space: O(n)
class Solution(object):
def longestUnivaluePath(self, root):
"""
:type root: TreeNode
:rtype: int
"""
longest = [0]
def traverse(node):
if not node:
return 0
left_len, right_len = traverse(node.left), traverse(node.right)
left = (left_len + 1) if node.left and node.left.val == node.val else 0
right = (right_len + 1) if node.right and node.right.val == node.val else 0
longest[0] = max(longest[0], left + right)
return max(left, right)
traverse(root)
return longest[0]
Python:
# Time: O(n)
# Space: O(h)
class Solution(object):
def longestUnivaluePath(self, root):
"""
:type root: TreeNode
:rtype: int
"""
result = [0]
def dfs(node):
if not node:
return 0
left, right = dfs(node.left), dfs(node.right)
left = (left+1) if node.left and node.left.val == node.val else 0
right = (right+1) if node.right and node.right.val == node.val else 0
result[0] = max(result[0], left+right)
return max(left, right) dfs(root)
return result[0]
C++:
class Solution {
public:
int longestUnivaluePath(TreeNode* root) {
int lup = 0;
if (root) dfs(root, lup);
return lup;
}
private:
int dfs(TreeNode* node, int& lup) {
int l = node->left ? dfs(node->left, lup) : 0;
int r = node->right ? dfs(node->right, lup) : 0;
int resl = node->left && node->left->val == node->val ? l + 1 : 0;
int resr = node->right && node->right->val == node->val ? r + 1 : 0;
lup = max(lup, resl + resr);
return max(resl, resr);
}
};
类似题目：
[LeetCode] 250. Count Univalue Subtrees 计数相同值子树的个数