leetcode面试准备:Count Complete Tree Nodes

1 题目

Given a complete binary tree, count the number of nodes.

In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

接口:public int countNodes(TreeNode root);

2 思路

思路1

用暴力法：不考虑完全二叉树的特性，直接遍历求节点数。---超时

复杂度: Time：O(N); Space：O(log N)

思路2

完全二叉树的节点数,可以用公式算出 2^h - 1. 如果高度不相等, 则递归调用 countNodes(root.left) + 1 + countNodes(root.right)

复杂度: Time：O(log N); Space：O(log N)

思路3

二分查找的思想，感觉和思路2差不多。但是代码写出来的效率高一些。

复杂度: Time：O(log N); Space：O(1)

3 代码

思路1

	public int countNodes0(TreeNode root) {
		if (root == null)
			return 0;
		return countNodes(root.left) + 1 + countNodes(root.right);
	}

思路2

	public int countNodes(TreeNode root) {
		if (root == null)
			return 0;
		int leftHigh = 0, rightHigh = 0;
		TreeNode lchild = root.left, rchild = root.right;
		for (; lchild != null;) {
			leftHigh++;
			lchild = lchild.left;
		}
		for (; rchild != null;) {
			rightHigh++;
			rchild = rchild.right;
		}

		if (leftHigh == rightHigh) {
			return (2 << leftHigh) - 1;
		} else {
			return countNodes(root.left) + 1 + countNodes(root.right);
		}
	}

4 总结

计算阶乘的注意： (2 << leftHigh) - 1才不会超时，用Math.pow()超时。

二分查找思想，自己很不熟练。多找题目联系联系。

5 参考

• leetcode
• leetcode 222: Count Complete Tree Nodes