代码随想录算法训练营20期|第十六天|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  

递归法：

/**
 * 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;

    }
}

迭代法

/**
 * 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 level = 0;
        Queue queue = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            level++;
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                if (cur.left != null) queue.add(cur.left);
                if (cur.right != null) queue.add(cur.right);
            }
        }
        return level;
    }
}

559.n叉树的最大深度

递归：

/*
// Definition for a Node.
class Node {
    public int val;
    public List children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public int maxDepth(Node root) {
        if (root == null) return 0;
        int depth = 0;
        for (Node child : root.children) {
            if (child != null) {
                depth = Math.max(depth, maxDepth(child));
            }
        }
        return depth + 1;
    }
}

迭代法

/*
// Definition for a Node.
class Node {
    public int val;
    public List children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public int maxDepth(Node root) {
        if (root == null) return 0;
        int depth = 0;
        Queue queue = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            depth++;
            for (int i = 0; i < size; i++) {
                Node cur = queue.poll();
                if (cur.children != null) {
                    for (Node child : cur.children) {
                        queue.add(child);
                    }
                }
            }
        }
        return depth;
    }
}

111.二叉树的最小深度

递归法

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

    }
}

迭代法

/**
 * 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;
        Queue queue = new LinkedList<>();
        queue.add(root);
        int depth = 0;

        while (!queue.isEmpty()) {
            int size = queue.size();
            depth++;

            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                if (cur.left == null && cur.right == null) {
                    return depth;
                }
                if (cur.left != null) queue.add(cur.left);
                if (cur.right != null) queue.add(cur.right);
            }
        }
        return depth;
    }
}

222.完全二叉树的节点个数

递归：

/**
 * 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);
        int res = left + right + 1;
        return res;

    }
}

迭代：

/**
 * 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 count = 0;
        Queue queue = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                count++;
                if (cur.left != null) queue.add(cur.left);
                if (cur.right != null) queue.add(cur.right);
            }
        }
        return count;
    }
}