662. Maximum Width of Binary Tree (DFS, BFS)
package LeetCode_662 import java.util.* import kotlin.collections.ArrayList /** * 662. Maximum Width of Binary Tree * https://leetcode.com/problems/maximum-width-of-binary-tree/description/ *https://www.cnblogs.com/grandyang/p/7538821.html * * Input: 1 / \ 3 2 / \ \ 5 3 9 Output: 4 Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9). * */ /** * Example: * var ti = TreeNode(5) * var v = ti.`val` * Definition for a binary tree node. * */ class TreeNode(var `val`: Int) { var left: TreeNode? = null var right: TreeNode? = null } /** * 所以这道题的关键就是要记录每一层中最左边结点的位置,我们知道对于一棵完美二叉树,如果根结点是深度1,那么每一层的结点数就是2*n-1, * 那么每个结点的位置就是[1, 2*n-1]中的一个,假设某个结点的位置是i,那么其左右子结点的位置可以直接算出来,为2*i和2*i+1,可以自行带例子检验。 * */ class Solution { /* * solution 1:dfs, Time:O(n), Space:O(h) * solution 2:bfs, Time:O(n), Space:O(n) * */ fun widthOfBinaryTree(root: TreeNode?): Int { //solution 1 /*val lefts = ArrayList<Int>() val res = IntArray(1) dfs(root, 0, 1, lefts, res) return res[0]*/ //solution 2: return bfs(root) } fun dfs(node: TreeNode?, depth: Int, id: Int, lefts: ArrayList<Int>, res: IntArray) { if (node == null) return if (depth >= lefts.size) lefts.add(id) res[0] = Math.max(res[0], id + 1 - lefts.get(depth)) //if the parent node is n, then the left child is 2*n, and right child is 2*n+1 dfs(node.left, depth + 1, id * 2, lefts, res) dfs(node.right, depth + 1, id * 2 + 1, lefts, res) } fun bfs(root: TreeNode?): Int { //queue for save node val queue = LinkedList<TreeNode>() //queue for save index val queueForIndex = LinkedList<Int>() var max = 0 queue.offer(root) //assuming root's index is 1 queueForIndex.offer(1) while (queue.isNotEmpty()) { //the size of current level val size = queue.size var start = 0 var end = 0 for (i in 0 until size) { val node = queue.poll() val curIndex = queueForIndex.poll() //the start, and of each level if (i == 0) { start = curIndex } if (i == size - 1) { end = curIndex } //if the parent node is n, then the left child is 2*n, and right child is 2*n+1 if (node.left != null) { queue.offer(node.left) queueForIndex.offer(2 * curIndex) } if (node.right != null) { queue.offer(node.right) queueForIndex.offer(2 * curIndex + 1) } } max = Math.max(max, end - start + 1) } return max } }