[leetcode]Count Complete Tree Nodes

题目描述如下：

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.

因为树这样的题目也没办法很好的调试，所以之前也是CE了好多遍

第一版的代码如下：

public class Solution {
public int res;
Solution(){
this.res = 0;
}
public void addCount(){
this.res++;
}
public int getCount(){
return this.res;
}
public int countNodes(TreeNode root) {
if(root == null) return 0;
addCount();
if(root.left == null && root.right == null)
return getCount();
if(root.left != null){
countNodes(root.left);
}
if(root.right != null)
countNodes(root.right);
return getCount();
}
}

然后就TLE了。想想也是，毕竟是medium难度的题目，如果就这样暴力破解未免太过分了。

再读了一遍题目，完全二叉树左子树高度和右子树高度一致，这时候若树的高度为height，那么一共有height^2-1个节点，这个地方可以进行剪枝。所以，只要递归左右两边子树高度不同的部分就好。利用这个思路再写就AC了。附上代码：

public class Solution {
public int res;
Solution(){
this.res = 0;
}
public int getLeftHeight(TreeNode root){
int height = 0;
while(root.left != null) {
root = root.left;
height ++;
}
return height;
}
public int getRightHeight(TreeNode root){
int height = 0;
while(root.right != null) {
root = root.right;
height ++;
}
return height;
}
public int countNodes(TreeNode root) {
if(root == null) return 0;
int lheight = getLeftHeight(root);
int rheight = getRightHeight(root);
if(lheight == rheight) return (2 << lheight) - 1;
else return countNodes(root.left) + countNodes(root.right) + 1;
}
}

题目链接：https://leetcode.com/problems/count-complete-tree-nodes/