[LeetCode] 687. Longest Univalue Path 最长唯一值路径

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　　

原文地址：https://www.cnblogs.com/lightwindy/p/9795761.html