[LeetCode-JAVA] Count Complete Tree Nodes

標籤：

題目：

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 2hnodes inclusive at the last level h.

題意：求一個完全二叉樹的節點個數。

思路：找到最後一層的最後一個節點，可以判斷左右節點最最左邊的層數是否相同，如果相同，則左子樹為滿二叉樹，若不同則右子樹為滿二叉樹。

其中在求解的過程中，需要用到冪次的運算，如果用Math.pow會逾時，可以考慮有位操作，但是要考慮2的0次冪的特殊情況。

代碼：

public class Solution {    public int countNodes(TreeNode root) {        if(root == null)            return 0;        int left = countLevel(root.left);        int right = countLevel(root.right);                int leftpow = 2<<(left-1);        int rightpow = 2<<(right-1);                if(left == 0)   //0次冪，<<不出來            leftpow = 1;        if(right == 0)            rightpow = 1;                if(left == right){            return leftpow + countNodes(root.right);        }else            return rightpow + countNodes(root.left);    }        public int countLevel(TreeNode root) {        if(root == null)            return 0;        int count = 0;        while(root != null) {            count++;            root = root.left;        }                return count;    }}

 

[LeetCode-JAVA] Count Complete Tree Nodes