【重点】【二叉树】662.二叉树最大宽度

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原创 2025-06-28 13:09:03 691 阅读 AI 写同款· GEO 优化 ›

算法学习（门徒计划）2-1 二叉树（Binary-Tree）与经典问题学习笔记学习二叉树的核心理念 - 数据的存储结构本身不具备绝对的优劣区分，最重要的是根据实际情况选择最合适的结构进行使用。 - 树结构是为了实现管理数据的从属关系和实现快速查询而设计的。 - 学习树有利于提升思维的维度。 - 设计方案时，时空复杂度有时不可兼得，需要根据具体情况选择合适的方式阅读详情

题目

法1：BFS，最佳方法

在这里插入图片描述

版本1：修改原TreeNode

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def widthOfBinaryTree(self, root: Optional[TreeNode]) -> int:
        res = 0
        queue = collections.deque()
        root.val = 1
        queue.append(root)
        while queue:
            tmpSize = len(queue)
            start, end = -1, -1
            for i in range(tmpSize):
                curNode = queue.popleft()
                val = curNode.val
                if i == 0:
                    start = curNode.val
                if i == tmpSize - 1:
                    end = curNode.val
                if curNode.left:
                    curNode.left.val = 2 * val
                    queue.append(curNode.left)
                if curNode.right:
                    curNode.right.val = 2 * val + 1
                    queue.append(curNode.right)
            res = max(res, end - start + 1)

        return res

Java

class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        int ans = 0;
        Deque<TreeNode> deque = new LinkedList<>();
        root.val = 1;
        deque.offer(root);
        while(!deque.isEmpty()) {
            int count = deque.size(), startIndex = -1, endIndex = -1;
            for (int i = 0; i < count; i++) {
                TreeNode node = deque.poll();
                endIndex = node.val;
                if (startIndex == -1) {
                    startIndex = node.val;
                }
                if (node.left != null) {
                    node.left.val = node.val * 2;
                    deque.offer(node.left);
                }
                if (node.right != null) {
                    node.right.val = node.val * 2 + 1;
                    deque.offer(node.right);
                }
            }
            ans = Math.max(ans, endIndex - startIndex + 1);
        }

        return ans;
    }
}

版本2：构建新TreeNode

class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        int ans = 0;
        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(new TreeNode(1, root.left, root.right));
        while(!deque.isEmpty()) {
            int count = deque.size(), startIndex = -1, endIndex = -1;
            for (int i = 0; i < count; i++) {
                TreeNode node = deque.poll();
                endIndex = node.val;
                if (startIndex == -1) {
                    startIndex = node.val;
                }
                if (node.left != null) {
                    deque.offer(new TreeNode(node.val * 2, node.left.left, node.left.right));
                }
                if (node.right != null) {
                    deque.offer(new TreeNode(node.val * 2 + 1, node.right.left, node.right.right));
                }
            }
            ans = Math.max(ans, endIndex - startIndex + 1);
        }

        return ans;
    }
}