1026. Maximum Difference Between Node and Ancestor
- Hardness: \(\color{orange}\textsf{Medium}\)
- Ralated Topics:
Tree、Depth-First Search、Binary Tree
一、題目
Given the root of a binary tree, find the maximum value v for which there exist different nodes a and b where v = |a.val - b.val| and a is an ancestor of b.
A node a is an ancestor of b if either: any child of a is equal to b or any child of a is an ancestor of b.
Example 1:
- Input: root = [8,3,10,1,6,null,14,null,null,4,7,13]
- Output: 7
- Explanation: We have various ancestor-node differences, some of which are given below :
|8 - 3| = 5
|3 - 7| = 4
|8 - 1| = 7
|10 - 13| = 3
Among all possible differences, the maximum value of 7 is obtained by |8 - 1| = 7.
Example 2:
- Input: root = [1,null,2,null,0,3]
- Output: 3
Constraints:
- The number of nodes in the tree is in the range
[2, 5000]. 0 <= Node.val <= 10^5
二、分析
- 借用一個
helper函式，將父節點的值往子葉節點傳，並且記錄過程中的最大值與最小值，並定義一個全域變數記錄其差值。
三、解題
1. DFS
- Time complexity: \(O(n)\)
- Space complexity: \(O(1)\)
int res = 0;
int maxAncestorDiff(TreeNode* root) {
helper(root, root->val, root->val);
return res;
}
void helper(TreeNode* root, int maxv, int minv) {
if (!root) return;
maxv = max(maxv, root->val);
minv = min(minv, root->val);
helper(root->left, maxv, minv);
helper(root->right, maxv, minv);
res = max(res, maxv-minv);
}