java计算二叉树的节点最小值

java计算二叉树的节点最小值

题目：

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 2h nodes

inclusive at the last level h.

解题：

假设用常规的解法一个个遍历，就是O(n)时间复杂度 。会不通过，这边就不写O(n)的代码了。由于是全然二叉树，满二叉树有一个性质是节点数等于2^h-1,h为高度，所以能够这样推断节点的左右高度是不是一样，假设是一样说明是满二叉树，就能够用刚才的公式。假设左右不相等就递归计算左右节点。

代码：

public static int countNodes(TreeNode root) {

if(root==null)

return 0;

else {

int left=getLeftHeight(root);

int right=getRightHeight(root);

if(left==right)

return (1<

else {

return countNodes(root.right)+countNodes(root.left)+1;

}

}

}

public static int getRightHeight(TreeNode root) {

int height=0;

while(root!=null)

{

height++;

root=root.left;

}

return height;

}

public static int getLeftHeight(TreeNode root) {

int height=0;

while(root!=null)

{

height++;

root=root.right;

}

return height;

}