1530. 好叶子节点对的数量

给你二叉树的根节点 root 和一个整数 distance 。

如果二叉树中两个 叶 节点之间的 最短路径长度 小于或者等于 distance ，那它们就可以构成一组 好叶子节点对 。

返回树中 好叶子节点对的数量 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-good-leaf-nodes-pairs
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {
    public int countPairs(TreeNode root, int distance) {
        Pair pair = dfs(root, distance);
        return pair.count;
    }

    public Pair dfs(TreeNode root, int distance) {
        int[] depths = new int[distance + 1];
        if (root == null) {
            return new Pair(depths, 0);
        }

        if (root.left == null && root.right == null) {
            depths[0] = 1;
            return new Pair(depths, 0);
        }

        Pair left = dfs(root.left, distance);
        Pair right = dfs(root.right, distance);

        for (int i = 0; i < distance; i++) {
            depths[i + 1] += left.depths[i] + right.depths[i];
        }

        int cnt = 0;
        for (int i = 0; i <= distance; i++) {
            for (int j = 0; j + i + 2 <= distance; j++) {
                cnt += left.depths[i] * right.depths[j];
            }
        }
        return new Pair(depths, cnt + left.count + right.count);
    }
}

class Pair {
    int[] depths;
    int count;

    public Pair(int[] depths, int count) {
        this.depths = depths;
        this.count = count;
    }
}

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;
    }
}

标签：distance,depths,int,dfs,叶子,1530,Pair,root,节点
来源： https://www.cnblogs.com/tianyiya/p/15840481.html