leetcode 662. Maximum Width of Binary Tree | 662. 二叉树最大宽度（BFS）

题目

https://leetcode.com/problems/maximum-width-of-binary-tree/

题解

本题思路来源于二叉树的层序遍历。

层序遍历类似问题：leetcode 199. Binary Tree Right Side View | 199. 二叉树的右视图（Java）
如何使用“记录下一行最右节点”的方式，进行层序遍历：左神算法：二叉树的按层打印与ZigZag打印

一开始就想到了给每一个元素维护 position，但误以为左孩子 pos-1，右孩子 pos+1，提交之后才发现并不是这样的。看了下答案的提示：

答案

这个问题中的主要想法是，利用完全二叉树（用数组存储堆）的性质，给每个节点一个 position 值，如果我们走向左子树，那么 position -> position * 2，如果我们走向右子树，那么 position -> positon * 2 + 1。当我们在看同一层深度的位置值 L 和 R 的时候，宽度就是 R - L + 1。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Info {TreeNode node;int index;public Info(TreeNode node, int index) {this.node = node;this.index = index;}
}class Solution {public int widthOfBinaryTree(TreeNode root) {Queue<Info> queue = new LinkedList<>();Info nextLast = new Info(root, 0);Info curFirst = nextLast;Info curLast = nextLast;queue.add(nextLast);int result = 0;while (queue.size() > 0) {Info poll = queue.poll();if (poll.node.left != null) {Info left = new Info(poll.node.left, poll.index * 2);queue.offer(left);nextLast = left;}if (poll.node.right != null) {Info right = new Info(poll.node.right, poll.index * 2 + 1);queue.offer(right);nextLast = right;}if (poll.node == curLast.node) { // after this node, goto a new lineresult = Math.max(result, curLast.index - curFirst.index + 1);curFirst = queue.peek();curLast = nextLast;}}return result;}
}

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leetcode 662. Maximum Width of Binary Tree | 662. 二叉树最大宽度（BFS）

题目

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答案

leetcode 662. Maximum Width of Binary Tree | 662. 二叉树最大宽度（BFS）相关推荐

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