代码随想录算法训练营-day20-104. 二叉树的最大深度、111. 二叉树的最小深度、559. N 叉树的最大深度、222. 完全二叉树的节点个数

104. 二叉树的最大深度

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

111. 二叉树的最小深度

/**
 * 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 depth = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            int size = queue.size();
            depth++;
            while(size-- >0){
                TreeNode temp = queue.poll();
                if(temp.left != null) queue.offer(temp.left);
                if(temp.right != null) queue.offer(temp.right);
                if(temp.left == null && temp.right == null){
                    return depth;
                    //遇到叶子节点直接返回即可，此时的深度就是最小深度
                }
            }
        }
        return depth;
    }

}

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

559. N 叉树的最大深度

/*
// 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 depth = 0;
        for (Node child : root.children) {
            int childDepth = maxDepth(child);
            depth = depth > childDepth ? depth : childDepth;
        }
        return depth + 1;
    }
}

222. 完全二叉树的节点个数

/**
 * 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 {
    //普通二叉树解法
    //迭代方法直接层序遍历的时候，记录下结点个数即可。
    //这里采用递归，与求解二叉树的最大深度，最小深度的方法类似。
    public int countNodes(TreeNode root) {
        if (root == null) return 0;
        int leftNum = countNodes(root.left);
        int rightNum = countNodes(root.right);
        int treeNum = leftNum + rightNum + 1;
        return treeNum;
    }
}

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

/**
 * 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 leftDepth = 0;		//初始化为0，为指数计算提供方便
        int rightDepth = 0;
        TreeNode left = root.left;
        TreeNode right = root.right;
        while (left != null) {
            left = left.left;
            leftDepth++;
        }
        while (right != null) {
            right = right.right;
            rightDepth++;
        }

        if (leftDepth == rightDepth) {
            return (2 << leftDepth) - 1;
        }
        return countNodes(root.left) + countNodes(root.right) + 1;
    }
}