LeetCode——Count Complete Tree Nodes

Description:

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 2hnodes inclusive at the last level h.

思路：给一颗完全二叉树求其节点的个数。首先要明确完全二叉树的定义：把满二叉树从上到下，从左到右进行编号，完全二叉树是其中编号没有断续的部分。也就是说完全二叉树只可能在最右子树的叶子节点的位子上有空缺。而且左右子树的高度差不能大于1。所以只要是count(左节点)==count(右节点)那么就是一颗满二叉树。可以用公式计算出节点的个数NodeCount=2^h - 1。若不是满二叉树的话就递归遍历求count(左子树) + count(右子树) + 1。

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