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152 lines
5.0 KiB
Markdown
Executable File
152 lines
5.0 KiB
Markdown
Executable File
# [662. Maximum Width of Binary Tree](https://leetcode.com/problems/maximum-width-of-binary-tree/)
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## 题目
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Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a **full binary tree**, but some nodes are null.
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The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the `null` nodes between the end-nodes are also counted into the length calculation.
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**Example 1**:
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Input:
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1
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/ \
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3 2
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/ \ \
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5 3 9
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Output: 4
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Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).
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**Example 2**:
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Input:
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1
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/
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3
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/ \
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5 3
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Output: 2
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Explanation: The maximum width existing in the third level with the length 2 (5,3).
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**Example 3**:
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Input:
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1
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/ \
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3 2
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/
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5
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Output: 2
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Explanation: The maximum width existing in the second level with the length 2 (3,2).
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**Example 4**:
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Input:
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1
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/ \
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3 2
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/ \
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5 9
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/ \
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6 7
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Output: 8
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Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).
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**Note**: Answer will in the range of 32-bit signed integer.
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## 题目大意
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给定一个二叉树，编写一个函数来获取这个树的最大宽度。树的宽度是所有层中的最大宽度。这个二叉树与满二叉树（full binary tree）结构相同，但一些节点为空。
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每一层的宽度被定义为两个端点（该层最左和最右的非空节点，两端点间的null节点也计入长度）之间的长度。
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注意: 答案在32位有符号整数的表示范围内。
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## 解题思路
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- 给出一个二叉树，求这棵树最宽的部分。
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- 这一题可以用 BFS 也可以用 DFS，但是用 BFS 比较方便。按照层序遍历，依次算出每层最左边不为 `null` 的节点和最右边不为 `null` 的节点。这两个节点之间都是算宽度的。最终输出最大的宽度即可。此题的关键在于如何有效的找到每一层的左右边界。
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- 这一题可能有人会想着先补全满二叉树，然后每层分别找左右边界。这种方法提交以后会卡在 `104 / 108` 这组测试用例上，这组测试用例会使得最后某几层填充出现的满二叉树节点特别多，最终导致 `Memory Limit Exceeded` 了。
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- 由于此题要找每层的左右边界，实际上每个节点的 `Val` 值是我们不关心的，那么可以把这个值用来标号，标记成该节点在每层中的序号。父亲节点在上一层中的序号是 x，那么它的左孩子在下一层满二叉树中的序号是 `2*x`，它的右孩子在下一层满二叉树中的序号是 `2*x + 1`。将所有节点都标上号，用 BFS 层序遍历每一层，每一层都找到左右边界，相减拿到宽度，动态维护最大宽度，就是本题的最终答案。
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## 代码
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```go
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package leetcode
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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func widthOfBinaryTree(root *TreeNode) int {
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if root == nil {
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return 0
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}
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if root.Left == nil && root.Right == nil {
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return 1
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}
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queue, res := []*TreeNode{}, 0
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queue = append(queue, &TreeNode{0, root.Left, root.Right})
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for len(queue) != 0 {
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var left, right *int
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// 这里需要注意，先保存 queue 的个数，相当于拿到此层的总个数
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qLen := len(queue)
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// 这里循环不要写 i < len(queue)，因为每次循环 queue 的长度都在变小
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for i := 0; i < qLen; i++ {
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node := queue[0]
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queue = queue[1:]
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if node.Left != nil {
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// 根据满二叉树父子节点的关系，得到下一层节点在本层的编号
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newVal := node.Val * 2
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queue = append(queue, &TreeNode{newVal, node.Left.Left, node.Left.Right})
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if left == nil || *left > newVal {
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left = &newVal
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}
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if right == nil || *right < newVal {
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right = &newVal
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}
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}
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if node.Right != nil {
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// 根据满二叉树父子节点的关系，得到下一层节点在本层的编号
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newVal := node.Val*2 + 1
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queue = append(queue, &TreeNode{newVal, node.Right.Left, node.Right.Right})
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if left == nil || *left > newVal {
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left = &newVal
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}
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if right == nil || *right < newVal {
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right = &newVal
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}
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}
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}
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switch {
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// 某层只有一个点，那么此层宽度为 1
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case left != nil && right == nil, left == nil && right != nil:
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res = max(res, 1)
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// 某层只有两个点，那么此层宽度为两点之间的距离
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case left != nil && right != nil:
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res = max(res, *right-*left+1)
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}
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}
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return res
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}
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``` |