LC.P1026[节点与其祖先之间的最大差值]

方法一：DFS

通过DFS，记录以该节点为根节点的路径上的最大值与最小值，最后更新答案。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
/**
 * 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 {
    int ans;

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

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