1026. Maximum Difference Between Node and Ancestor (M)

Maximum Difference Between Node and Ancestor (M)

题目

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

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题意

找到一组结点，其中一个是另一个祖先结点，使得这两个结点的差值最大。

思路

递归处理，两种方法：

1. 每次找到左子树和右子树中的最值，以此更新结果。
2. 每次递归更新当前路径上的最大值和最小值，到达叶结点时计算差值并返回。

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代码实现

Java

递归1

class Solution {
    private int diff;

    public int maxAncestorDiff(TreeNode root) {
        diff = 0;
        dfs(root);
        return diff;
    }

    private int[] dfs(TreeNode root) {
        if (root == null) {
            return null;
        }

        int[] l = dfs(root.left), r = dfs(root.right);

        if (l == null && r == null) {
            return new int[] { root.val, root.val };
        }

        int cMin = 0, cMax = 0;
        if (l != null && r != null) {
            cMin = Math.min(l[0], r[0]);
            cMax = Math.max(l[1], r[1]);
        } else if (l != null) {
            cMin = l[0];
            cMax = l[1];
        } else {
            cMin = r[0];
            cMax = r[1];
        }
        diff = Math.max(diff, Math.max(Math.abs(root.val - cMin), Math.abs(root.val - cMax)));
        return new int[] { Math.min(root.val, cMin), Math.max(root.val, cMax) };
    }
}

递归2

class Solution {
    public int maxAncestorDiff(TreeNode root) {
        if (root == null) {
            return 0;
        }

        return dfs(root, root.val, root.val);
    }

    private int dfs(TreeNode root, int max, int min) {
        if (root == null) {
            return max - min;
        }

        max = Math.max(root.val, max);
        min = Math.min(root.val, min);
        return Math.max(dfs(root.left, max, min), dfs(root.right, max, min));
    }
}