LC.P1448[统计二叉树中好节点的数目]

方法一：DFS

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/**
 * 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 {
    
    public int goodNodes(TreeNode root) {
        return dfs(root, Integer.MIN_VALUE);
    }

    private int dfs(TreeNode root, int max) {
        if (root == null) return 0;
        int left = dfs(root.left, Math.max(max, root.val));
        int right = dfs(root.right, Math.max(max, root.val));
        return left + right + (max <= root.val ? 1 : 0);
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$