1026. Maximum Difference Between Node and Ancestor

# 1026. Maximum Difference Between Node and Ancestor#

## 题目 #

Given the `root` of a binary tree, find the maximum value `V` for which there exists 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: [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.
``````

Note:

1. The number of nodes in the tree is between `2` and `5000`.
2. Each node will have value between `0` and `100000`.

## 题目大意 #

• 树中的节点数在 2 到 5000 之间。
• 每个节点的值介于 0 到 100000 之间。

## 解题思路 #

• 给出一颗树，要求找出祖先和孩子的最大差值。
• DPS 深搜即可。每个节点和其所有孩子的`最大值`来自于 3 个值，节点本身，递归遍历左子树的最大值，递归遍历右子树的最大值；每个节点和其所有孩子的`最小值`来自于 3 个值，节点本身，递归遍历左子树的最小值，递归遍历右子树的最小值。依次求出自身节点和其所有孩子节点的最大差值，深搜的过程中动态维护最大差值即可。

## 代码 #

``````func maxAncestorDiff(root *TreeNode) int {
res := 0
dfsAncestorDiff(root, &res)
return res
}

func dfsAncestorDiff(root *TreeNode, res *int) (int, int) {
if root == nil {
return -1, -1
}
leftMax, leftMin := dfsAncestorDiff(root.Left, res)
if leftMax == -1 && leftMin == -1 {
leftMax = root.Val
leftMin = root.Val
}
rightMax, rightMin := dfsAncestorDiff(root.Right, res)
if rightMax == -1 && rightMin == -1 {
rightMax = root.Val
rightMin = root.Val
}
*res = max(*res, max(abs(root.Val-min(leftMin, rightMin)), abs(root.Val-max(leftMax, rightMax))))
return max(leftMax, max(rightMax, root.Val)), min(leftMin, min(rightMin, root.Val))
}
``````

Apr 8, 2023