Count Complete Tree Nodes

Count Complete Tree Nodes

问题：

Given 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.

思路：

　　分治法的应用

他人代码：

public class Solution {

    public int countNodes(TreeNode root) {

        if(root == null)    return 0;

        int h = getHeight(root);

        return getHeight(root.right)==h-1 ? (1<<(h-1)) + countNodes(root.right) : (1<<(h-2)) + +countNodes(root.left);

    }

    public int getHeight(TreeNode root)

    {

        return root==null ? 0 : 1+getHeight(root.left);

    }

}

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• 关键之处在于通过获取高度确定，左子树满了还是没有满，若满了则只需要求解右子树的节点数就可以了，若未满，则需要求解左子树的节点数，典型的分治思想，主要一点在于求解高度进化划分问题成子问题。
• 最近写代码不想思考啊，这道题本可以自己最初来的，看了别人的代码，浪费了一道这么好的题目。Since I halve the tree in every recursive step, I have O(log(n)) steps. Finding a height costs O(log(n)). So overall O(log(n)^2).
• 改掉不好的习惯，最近的习惯有点不好，坏习惯有点多。