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1026. Maximum Difference Between Node and Ancestor

Posted on Edited on In Disqus:
Symbols count in article: 1.1k Reading time ≈ 1 mins.

O(n) time O(h) space
每次更新mx和mn，到nullptr时得到根到叶的一条路径上的最大和最小值，只需要求差即可，对于左右两个子树返回上来的不同路径的最大最小值之差，只需要取较大的那个即可

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxAncestorDiff(TreeNode* root) {
        return dfs(root, root->val, root->val);
    }

    int dfs(TreeNode *root, int mx, int mn) { // 输入从根到当前节点之前的最大和最小值
        return root ? max(dfs(root->left, max(mx, root->val), min(mn, root->val)), dfs(root->right, max(mx, root->val), min(mn, root->val))) : mx - mn;
    }
};
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxAncestorDiff(TreeNode* root) {
        return dfs(root, root->val, root->val); // 初始最大最小都为root即可
    }

    int dfs(TreeNode *root, int mx, int mn) { // 因为是绝对值，所以要维护最大最小两个值
        if (!root) return mx - mn;
        mx = max(mx, root->val);
        mn = min(mn, root->val);
        return max(dfs(root->left, mx, mn), dfs(root->right, mx, mn));
    }
};