1026. Maximum Difference Between Node and Ancestor
Brute force
- choose every node as an ancestor
- apply DFS to find maximum difference.
Implementation
```cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int ans = 0;
void preOrder(TreeNode* root, int val) {
if (root == nullptr) return;
ans = max(ans, abs(val - root -> val));
preOrder(root -> left, val);
preOrder(root -> right, val);
}
int maxAncestorDiff(TreeNode* root) {
queue<TreeNode*> qu;
qu.push(root);
while (!qu.empty()) {
auto u = qu.front(); qu.pop();
preOrder(u, u -> val);
if (u->left) qu.push(u -> left);
if (u->right) qu.push(u -> right);
}
return ans;
}
};
```
optimal
- DFS it
- meanwhile keep track of maximum and minimum values encounter.
- make sure to pass them as parameter, because they are unique to every node.
Implementation
```cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int ans;
void pre_order(TreeNode* root, int mx, int mn) {
if (root == nullptr) return;
mx = max(mx, root -> val);
mn = min(mn, root -> val);
ans = max(ans, abs(mx - mn));
pre_order(root->left, mx, mn);
pre_order(root->right, mx, mn);
}
int maxAncestorDiff(TreeNode* root) {
ans = 0;
pre_order(root, root->val, root->val);
return ans;
}
};
```