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package byteDance;
/**
* Created by lizeyang on 2020/5/16.
* 求完全二叉树的节点个数，小于O(n)，并分析复杂度
*/
public class test49 {
//O(n)，空间复杂度O(d)=O(logn),d是二叉树的高度
public static int test(TreeNode root) {
return root != null ? 1 + test(root.left) + test(root.right) : 0;
}
//利用完全二叉树的特点，除了最后一层外，其余每层节点都是满的，并且最后一层的节点全部靠向左边
// 二分搜索
public static int countNodes(TreeNode root) {
if (root == null) {
return 0;
}
int d = getDepth(root);
if (d == 0) {
return 1;
}
int l = 1, r = (int) Math.pow(2, d) - 1;
int mid;
while (l <= r) {
mid = (l + r) >> 1;
if (exist(mid, d, root)) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return (int) Math.pow(2, d) - 1 + l;
}
public static boolean exist(int cur, int level, TreeNode root) {
int l = 0, r = (int) Math.pow(2, level) - 1;
int mid;
for (int i = 0; i < level; i++) {
mid = (l + r) >> 1;
if (cur <= mid) {
root = root.left;
r = mid;
} else {
root = root.right;
l = mid + 1;
}
}
return root != null;
}
public static int getDepth(TreeNode root) {
if (root == null) {
return 0;
}
int level = 0;
while (root != null) {
level++;
root = root.left;
}
return level;
}
public static void main(String[] args) {
TreeNode root = new TreeNode(5);
root.left = new TreeNode(4);
root.right = new TreeNode(8);
root.left.left = new TreeNode(3);
root.left.right = new TreeNode(6);
root.right.left = new TreeNode(7);
root.right.right = new TreeNode(9);
root.left.left.left = new TreeNode(10);
//O(n)
System.out.println(test(root));
//O(logn)
System.out.println(countNodes(root));
}
}