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.
[分析]
[思路1]直接递归时间复杂度是2^h 级别,超时,需要些剪枝技巧:如果从当前节点一直往左的高度 == 一直往右的高度,则以当前节点为根的树是一颗perfect tree,包含的节点数 = 2^h - 1, 其中h 为根节点高度记为1时的树高;否则递归调用count(root.left) + count(root.right) + 1 来计算。剪枝后算法复杂度为h^2,h 为数高,可以这样理解:第一次检查左右子树高度时,高度为h,计算量时O(h),若有第二次,待检查的树高减一,计算量为O(h - 1)……如此递推,因此平均时间复杂度是O(h^2)。
[思路2] 二分枚举。高度为 h 的完全二叉树,其节点数=高度为 h - 1的满二叉树 + 最后一层节点数。如何获取最后一层节点数呢?h 从0开始计,则最后一层满时节点数=2^h。给最后一层几点编号,若记往左走为0,往右走为1,则最后一层节点编号二进制形式正好对应从根节点开始的路径。二分枚举节点并判断其是否存在,直到找到最后一层的最右边节点。
搬一个参考博客上的图:
Lv0 1
/ \
Lv1 2 3
/ \ / \
Lv2 4 5 6 -
No. 00 01 10
[ref]
http://bookshadow.com/weblog/2015/06/06/leetcode-count-complete-tree-nodes/
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { // Method 1: O(h^2) public int countNodes1(TreeNode root) { if (root == null) return 0; if (root.left == null && root.right == null) return 1; TreeNode curr = root; int lHeight = 0; while (curr.left != null) { lHeight++; curr = curr.left; } curr = root; int rHeight = 0; while (curr.right != null) { rHeight++; curr = curr.right; } if (lHeight == rHeight) { return (1 << (lHeight + 1)) - 1; } else { return countNodes(root.left) + countNodes(root.right) + 1; } } // Method 2: Binary Search public int countNodes(TreeNode root) { if (root == null) return 0; int height = 0; TreeNode curr = root; while (curr.left != null) { height++; curr = curr.left; } int left = 0, right = (1 << height) - 1; while (left <= right) { int mid = left + ((right - left) >> 1); if (existsNode(root, mid, height - 1)) { left = mid + 1; } else { right = mid - 1; } } return (1 << height) + right; } public boolean existsNode(TreeNode root, int path, int height) { while (height >= 0 && root != null) { if (((path >> height) & 1) == 1) { root = root.right; } else { root = root.left; } height--; } return root != null; } }