222. 完全二叉树的节点个数
给出一个完全二叉树,求出该树的节点个数。
说明:
完全二叉树的定义如下:在完全二叉树中,除了最底层节点可能没填满外,其余每层节点数都达到最大值,并且最下面一层的节点都集中在该层最左边的若干位置。若最底层为第 h 层,则该层包含 1~ 2h 个节点。
示例:
输入:
1
/ \
2 3
/ \ /
4 5 6
输出: 6
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
// Return tree depth in O(d) time.
public int computeDepth(TreeNode node) {
int d = 0;
while (node.left != null) {
node = node.left;
++d;
}
return d;
}
// Last level nodes are enumerated from 0 to 2**d - 1 (left -> right).
// Return True if last level node idx exists.
// Binary search with O(d) complexity.
public boolean exists(int idx, int d, TreeNode node) {
int left = 0, right = (int)Math.pow(2, d) - 1;
int pivot;
for(int i = 0; i < d; ++i) {
pivot = left + (right - left) / 2;
if (idx <= pivot) {
node = node.left;
right = pivot;
}
else {
node = node.right;
left = pivot + 1;
}
}
return node != null;
}
public int countNodes(TreeNode root) {
// if the tree is empty
if (root == null) return 0;
int d = computeDepth(root);
// if the tree contains 1 node
if (d == 0) return 1;
// Last level nodes are enumerated from 0 to 2**d - 1 (left -> right).
// Perform binary search to check how many nodes exist.
int left = 1, right = (int)Math.pow(2, d) - 1;
int pivot;
while (left <= right) {
pivot = left + (right - left) / 2;
if (exists(pivot, d, root)) left = pivot + 1;
else right = pivot - 1;
}
// The tree contains 2**d - 1 nodes on the first (d - 1) levels
// and left nodes on the last level.
return (int)Math.pow(2, d) - 1 + left;
}
}
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