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62 lines
2.1 KiB
Markdown
62 lines
2.1 KiB
Markdown
# [872. Leaf-Similar Trees](https://leetcode.com/problems/leaf-similar-trees/)
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## 题目
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Consider all the leaves of a binary tree. From left to right order, the values of those leaves form a *leaf value sequence.*
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For example, in the given tree above, the leaf value sequence is `(6, 7, 4, 9, 8)`.
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Two binary trees are considered *leaf-similar* if their leaf value sequence is the same.
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Return `true` if and only if the two given trees with head nodes `root1` and `root2` are leaf-similar.
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**Note**:
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- Both of the given trees will have between `1` and `100` nodes.
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## 题目大意
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请考虑一颗二叉树上所有的叶子，这些叶子的值按从左到右的顺序排列形成一个 叶值序列 。举个例子，如上图所示，给定一颗叶值序列为 (6, 7, 4, 9, 8) 的树。如果有两颗二叉树的叶值序列是相同，那么我们就认为它们是 叶相似 的。如果给定的两个头结点分别为 root1 和 root2 的树是叶相似的，则返回 true；否则返回 false 。
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提示：
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- 给定的两颗树可能会有 1 到 200 个结点。
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- 给定的两颗树上的值介于 0 到 200 之间。
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## 解题思路
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- 给出 2 棵树，如果 2 棵树的叶子节点组成的数组是完全一样的，那么就认为这 2 棵树是“叶子相似”的。给出任何 2 棵树判断这 2 棵树是否是“叶子相似”的。
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- 简单题，分别 DFS 遍历 2 棵树，把叶子节点都遍历出来，然后分别比较叶子节点组成的数组是否完全一致即可。
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## 代码
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```go
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func leafSimilar(root1 *TreeNode, root2 *TreeNode) bool {
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leaf1, leaf2 := []int{}, []int{}
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dfsLeaf(root1, &leaf1)
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dfsLeaf(root2, &leaf2)
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if len(leaf1) != len(leaf2) {
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return false
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}
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for i := range leaf1 {
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if leaf1[i] != leaf2[i] {
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return false
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}
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}
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return true
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}
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func dfsLeaf(root *TreeNode, leaf *[]int) {
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if root != nil {
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if root.Left == nil && root.Right == nil {
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*leaf = append(*leaf, root.Val)
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}
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dfsLeaf(root.Left, leaf)
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dfsLeaf(root.Right, leaf)
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}
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}
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``` |