[LeetCode] 687. Longest Univalue Path_Easy tag: DFS recursive

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.

这个题目的思路实际上跟 [LeetCode] 124. Binary Tree Maximum Path Sum 很像, 就是用一个helper function(input: root, output: 包括root的最长的univalue path 的number of nodes),

然后注意self.ans 是跟 1+ l + r 比较, 但是返回的时候只返回 1+ max(l, r), 最后返回self.ans -1 (因为题目求的是边)

1. Constraints

1) empty => 0

2) 1 node => 0

3) element will be interger

2. Ideas

DFS      T: O(n)     S; O(n)

1) self.ans = 1 # 这样方便判断如果not root 的情况

2) helper function, check l, r , if not 0, check root.val == root.child.val, update self.ans = max(self.ans, 1+ l+ r)

3) return self.ans -1

3. Code

class Solution:
    def maxUnivaluePath(self, root):
        ans = [None]
        def maxPathWithRoot(root):
            if not root: return 0
            left, right = maxPathWithRoot(root.left), maxPathWithRoot(root.right)
            left = 1 + left if root.left and root.val == root.left.val else 0
            right = 1 + right if root.right and root.val == root.right.val else 0
            local = max(0, left, right)
            potential = max(local, left + right)
            if ans[0] is None or potential > ans[0]:
                ans[0] = potential
            return local
        if not root: return 0
        maxPathWithRoot(root)
        return ans[0]