DAY16——二叉树part3

1.求二叉树深度

使用递归法完成

/**
 * 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;
        }
        return Math.max(maxDepth(root.left),maxDepth(root.right))+1;
    }
}

n叉树深度

/*
// 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 max = 0;
        for(Node node: root.children){
            max = Math.max(maxDepth(node), max);
        }
        return max + 1;
    }
}

需要注意题目给出的数据接口，这个题中为children

2.求二叉树的最小深度

需要注意最小深度需要是叶子节点

/*
// 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 max = 0;
        for(Node node: root.children){
            max = Math.max(maxDepth(node), max);
        }
        return max + 1;
    }
}

3.求完全二叉树的节点数

class Solution {
    public int countNodes(TreeNode root) {
        if(root == null){
            return 0;
        }
        return countNodes(root.left)+countNodes(root.right)+1;
    }
}

class Solution {
    // 迭代法
    public int countNodes(TreeNode root) {
        if (root == null) return 0;
        Queue queue = new LinkedList<>();
        queue.offer(root);
        int result = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            while (size > 0) {
                TreeNode cur = queue.poll();
                result++;
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
                size--;
            }
        }
        return result;
    }
}