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_1104.java
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package com.fishercoder.solutions;
import com.fishercoder.common.classes.TreeNode;
import com.fishercoder.common.utils.CommonUtils;
import com.fishercoder.common.utils.TreeUtils;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Deque;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
/**
* 1104. Path In Zigzag Labelled Binary Tree
*
* In an infinite binary tree where every node has two children, the nodes are labelled in row order.
* In the odd numbered rows (ie., the first, third, fifth,...), the labelling is left to right,
* while in the even numbered rows (second, fourth, sixth,...), the labelling is right to left.
*
* Given the label of a node in this tree, return the labels in the path from the root of the tree to the node with that label.
*
* Example 1:
* Input: label = 14
* Output: [1,3,4,14]
*
* Example 2:
* Input: label = 26
* Output: [1,2,6,10,26]
*
* Constraints:
* 1 <= label <= 10^6
* */
public class _1104 {
public static class Solution1 {
/**This brute force solution is correct but results in TLE on LeetCode.*/
public List<Integer> pathInZigZagTree(int label) {
Deque<Integer> deque = buildZigZagOrderList(label);
CommonUtils.printDeque(deque);
TreeNode root = buildZigZagOrderTree(deque);
TreeUtils.printBinaryTree(root);
return dfs(root, label, new ArrayList<>());
}
private List<Integer> dfs(TreeNode root, int label, List<Integer> list) {
if (root == null) {
return list;
}
list.add(root.val);
if (root.val == label) {
return list;
}
dfs(root.left, label, list);
dfs(root.right, label, list);
if (list.get(list.size() - 1) == label) {
return list;
}
list.remove(list.size() - 1);
return list;
}
private TreeNode buildZigZagOrderTree(Deque<Integer> deque) {
TreeNode root = new TreeNode(deque.pollFirst());
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!deque.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode curr = queue.poll();
curr.left = new TreeNode(deque.pollFirst());
curr.right = new TreeNode(deque.pollFirst());
queue.offer(curr.left);
queue.offer(curr.right);
}
}
return root;
}
private Deque<Integer> buildZigZagOrderList(int label) {
Deque<Integer> deque = new LinkedList<>();
int num = 1;
int level = 2;
deque.add(num);
do {
num++;
List<Integer> newLevel = new ArrayList<>();
for (; num < Math.pow(2, level); num++) {
newLevel.add(num);
}
num--;
if (level % 2 == 0) {
Collections.reverse(newLevel);
}
deque.addAll(newLevel);
newLevel.clear();
level++;
} while (deque.getLast() < label);
return deque;
}
}
public static class Solution2 {
/**We'll directly compute the index of its parent, it'll be much faster this way.*/
public List<Integer> pathInZigZagTree(int label) {
List<List<Integer>> lists = buildZigZagOrderList(label);
List<Integer> result = new ArrayList<>();
int index = findIndex(lists.get(lists.size() - 1), label);
result.add(label);
for (int i = lists.size() - 2; i >= 0; i--) {
index /= 2;
result.add(lists.get(i).get(index));
}
Collections.sort(result);
return result;
}
private int findIndex(List<Integer> level, int label) {
for (int i = 0; i < level.size(); i++) {
if (level.get(i) == label) {
return i;
}
}
return -1;
}
private List<List<Integer>> buildZigZagOrderList(int label) {
List<List<Integer>> lists = new ArrayList<>();
int num = 1;
int level = 2;
lists.add(Arrays.asList(num));
if (label == 1) {
return lists;
}
List<Integer> newLevel = new ArrayList<>();
do {
newLevel.clear();
num++;
for (; num < Math.pow(2, level); num++) {
newLevel.add(num);
}
num--;
if (level % 2 == 0) {
Collections.reverse(newLevel);
}
lists.add(new ArrayList<>(newLevel));
level++;
} while (newLevel.get(0) < label && newLevel.get(newLevel.size() - 1) < label);
return lists;
}
}
}