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CountCompleteTreeNodes.java
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CountCompleteTreeNodes.java
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package binarysearch;
// Source : https://leetcode.com/problems/count-complete-tree-nodes/
// Id : 222
// Author : Fanlu Hai | https://github.com/Fanlu91/FanluLeetcode
// Date : 2020/8/17
// Topic : Binary Search
// Level : Medium
// Other :
// Tips :
// Links :
// Result : 100.00% 53.27%
public class CountCompleteTreeNodes {
public int countNodes(TreeNode root) {
return root != null ? 1 + countNodes(root.right) + countNodes(root.left) : 0;
}
// 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 countNodes1(TreeNode root) {
// 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;
}
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
}