统计唯一值子树个数——lintcode921/leetcode250

统计唯一值子树个数

题目：统计唯一值子树个数

给定一棵二叉树，统计唯一值子树的数目.唯一值子树意味着子树的所有节点都具有相同的值.

示例：

输入: root = {5,1,5,5,5,#,5}
输出: 4
解释:
              5
             / \
            1   5
           / \   \
          5   5   5

题解

1. BFS+DFS

BFS遍历每个节点，DFS判断每个节点是否正确。(时间效率不高，懒人法)

public class Solution {
    private int sum;

    //dfs判断是否是同值子树
    public boolean dfs(TreeNode root, int key) {
        if (root == null) return true;
        if (root.val != key) return false;
        return dfs(root.left, key) && dfs(root.right, key);
    }

    public int countUnivalSubtrees(TreeNode root) {
        sum = 0;
        if (root == null) return sum;
        Queue<TreeNode> queue = new ArrayDeque<>();
        queue.add(root);

        //bfs遍历二叉树
        while (!queue.isEmpty()) {
            TreeNode temp = queue.poll();
            if (dfs(temp, temp.val)) sum++;
            if (temp.left != null) queue.add(temp.left);
            if (temp.right != null) queue.add(temp.right);
        }
        return sum;
    }
}

2. DFS判断每个节点

public class Solution2 {
    private int sum;

    public boolean dfs(TreeNode root) {
        if (root == null) return true;
        boolean left = dfs(root.left);
        boolean right = dfs(root.right);
        if (left && right) {
            if ((root.left != null && root.left.val != root.val) || (root.right != null && root.right.val != root.val)) {
                return false;
            }
            sum++;
            return true;
        }
        return false;
    }

    public int countUnivalSubtrees(TreeNode root) {
        sum = 0;
        dfs(root);
        return sum;
    }
}