222. 完全二叉树的节点个数 #

Difficulty: 中等

给出一个完全二叉树，求出该树的节点个数。

说明：

的定义如下：在完全二叉树中，除了最底层节点可能没填满外，其余每层节点数都达到最大值，并且最下面一层的节点都集中在该层最左边的若干位置。若最底层为第 h 层，则该层包含 1~ 2^h 个节点。

示例：

输入: 
    1
   / \
  2   3
 / \  /
4  5 6

输出: 6

题解 #

解法一：递归遍历（前、中、后序遍历） #

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        List<TreeNode> list = new ArrayList<>();
        helper(root, list);
        return list.size();
    }

    private void helper(TreeNode node, List<TreeNode> list) {
        if (node == null) {
            return;
        }
        list.add(node);
        helper(node.left, list);
        helper(node.right, list);
    }
}

解法二：递归遍历（一行递归） #

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return countNodes(root.left) + countNodes(root.right) + 1;
    }
}

解法三：二分查找 + 位运算（官方） #

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int level = 0;
        TreeNode node = root;
        while (node.left != null) {
            level++;
            node = node.left;
        }
        int low = 1 << level, high = (1 << (level + 1)) - 1;
        while (low < high) {
            int mid = (high - low + 1) / 2 + low;
            if (exists(root, level, mid)) {
                low = mid;
            } else {
                high = mid - 1;
            }
        }
        return low;
    }

    private boolean exists(TreeNode root, int level, int k) {
        int bits = 1 << (level - 1);
        TreeNode node = root;
        while (node != null && bits > 0) {
            if ((bits & k) == 0) {
                node = node.left;
            } else {
                node = node.right;
            }
            bits >>= 1;
        }
        return node != null;
    }
}