222. Count Complete Tree Nodes

https://leetcode.com/problems/count-complete-tree-nodes/#/description

http://www.cnblogs.com/EdwardLiu/p/5058570.html

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

Definition of a complete binary tree from Wikipedia:
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.

递归树高法

复杂度

时间 O(N) 空间 O(1)

思路

完全二叉树的一个性质是，如果左子树最左边的深度，等于右子树最右边的深度，说明这个二叉树是满的，即最后一层也是满的，则以该节点为根的树其节点一共有2^h-1个。如果不等于，则是左子树的节点数，加上右子树的节点数，加上自身这一个。

注意

• 这里在左节点递归时代入了上次计算的左子树最左深度减1，右节点递归的时候代入了上次计算的右子树最右深度减1，可以避免重复计算这些深度

• 做2的幂时不要用Math.pow，这样会超时。用1<<height这个方法来得到2的幂

/**
 * 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;
        int leftHeight = countLeft(root);
        int rightHeight = countRight(root);
        if (leftHeight == rightHeight) { //perfect binary tree
            return (1<<leftHeight)-1;
        }
        else return countNodes(root.left) + countNodes(root.right) + 1;
    }
    
    public int countLeft(TreeNode root) {
        int res = 0;
        while (root != null) {
            res++;
            root = root.left;
        }
        return res;
    }
    
    public int countRight(TreeNode root) {
        int res = 0;
        while (root != null) {
            res++;
            root = root.right;
        }
        return res;
    }
    
}