代码随想录训练营第三十期|第十六天|二叉树part03|104.二叉树的最大深度 559.n叉树的最大深度● 111.二叉树的最小深度● 222.完全二叉树的节点个数

104. 二叉树的最大深度 - 力扣（LeetCode）

/**
 * 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 maxDepth(TreeNode root) {
        if (root == null) return 0;
        int left = maxDepth(root.left);
        int right =maxDepth(root.right);
        return Math.max(left, right) + 1;
    }
}

111. 二叉树的最小深度 - 力扣（LeetCode）

/**
 * 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 minDepth(TreeNode root) {
        if (root == null) return 0;
        int left = minDepth(root.left);
        int right = minDepth(root.right);
        if (root.left == null && root.right != null) return right + 1;
        if (root.right == null && root.left != null) return left + 1;
        return Math.min(left, right) + 1;
    }
}

222. 完全二叉树的节点个数 - 力扣（LeetCode）

/**
 * 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 countNodes(TreeNode root) {
        if (root == null) return 0;
        int left = countNodes(root.left);
        int right = countNodes(root.right);
        return left + right + 1;
    }
}