# 1026. Maximum Difference Between Node and Ancestor

## Description

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 <= 105

## Solutions

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int ans;

public int maxAncestorDiff(TreeNode root) {
dfs(root, root.val, root.val);
return ans;
}

private void dfs(TreeNode root, int mi, int mx) {
if (root == null) {
return;
}
int x = Math.max(Math.abs(mi - root.val), Math.abs(mx - root.val));
ans = Math.max(ans, x);
mi = Math.min(mi, root.val);
mx = Math.max(mx, root.val);
dfs(root.left, mi, mx);
dfs(root.right, mi, mx);
}
}

• /**
* 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 maxAncestorDiff(TreeNode* root) {
int ans = 0;
function<void(TreeNode*, int, int)> dfs = [&](TreeNode* root, int mi, int mx) {
if (!root) {
return;
}
ans = max({ans, abs(mi - root->val), abs(mx - root->val)});
mi = min(mi, root->val);
mx = max(mx, root->val);
dfs(root->left, mi, mx);
dfs(root->right, mi, mx);
};
dfs(root, root->val, root->val);
return ans;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def maxAncestorDiff(self, root: Optional[TreeNode]) -> int:
def dfs(root, mi, mx):
if root is None:
return
nonlocal ans
ans = max(ans, abs(mi - root.val), abs(mx - root.val))
mi = min(mi, root.val)
mx = max(mx, root.val)
dfs(root.left, mi, mx)
dfs(root.right, mi, mx)

ans = 0
dfs(root, root.val, root.val)
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func maxAncestorDiff(root *TreeNode) (ans int) {
var dfs func(*TreeNode, int, int)
dfs = func(root *TreeNode, mi, mx int) {
if root == nil {
return
}
ans = max(ans, max(abs(mi-root.Val), abs(mx-root.Val)))
mi = min(mi, root.Val)
mx = max(mx, root.Val)
dfs(root.Left, mi, mx)
dfs(root.Right, mi, mx)
}
dfs(root, root.Val, root.Val)
return
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function maxAncestorDiff(root: TreeNode | null): number {
const dfs = (root: TreeNode | null, mi: number, mx: number): void => {
if (!root) {
return;
}
ans = Math.max(ans, Math.abs(root.val - mi), Math.abs(root.val - mx));
mi = Math.min(mi, root.val);
mx = Math.max(mx, root.val);
dfs(root.left, mi, mx);
dfs(root.right, mi, mx);
};
let ans: number = 0;
dfs(root, root.val, root.val);
return ans;
}


• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var maxAncestorDiff = function (root) {
let ans = 0;
const dfs = (root, mi, mx) => {
if (!root) {
return;
}
ans = Math.max(ans, Math.abs(mi - root.val), Math.abs(mx - root.val));
mi = Math.min(mi, root.val);
mx = Math.max(mx, root.val);
dfs(root.left, mi, mx);
dfs(root.right, mi, mx);
};
dfs(root, root.val, root.val);
return ans;
};


• /**
* Definition for a binary tree node.
* public class TreeNode {
*     public int val;
*     public TreeNode left;
*     public TreeNode right;
*     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
public class Solution {
private int ans;

public int MaxAncestorDiff(TreeNode root) {
dfs(root, root.val, root.val);
return ans;
}

private void dfs(TreeNode root, int mi, int mx) {
if (root == null) {
return;
}
int x = Math.Max(Math.Abs(mi - root.val), Math.Abs(mx - root.val));
ans = Math.Max(ans, x);
mi = Math.Min(mi, root.val);
mx = Math.Max(mx, root.val);
dfs(root.left, mi, mx);
dfs(root.right, mi, mx);
}
}