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# 222.Count-Complete-Tree-Nodes

## 题目描述

Given a complete binary tree, count the number of nodes.
Note:
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.
Example:
Input:
1
/ \
2 3
/ \ /
4 5 6
Output: 6

## 代码

### Approach #1 Linear Time

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int countNodes(TreeNode root) {
return root != null ? 1 + countNodes(root.right) + countNodes(left) : 0;
}
}
class Solution {
// 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;
}
// 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;
}
}