0872. Leaf Similar Trees

# 872. Leaf-Similar Trees#

## 题目 #

Consider all the leaves of a binary tree. From left to right order, the values of those leaves form a leaf value sequence.

For example, in the given tree above, the leaf value sequence is `(6, 7, 4, 9, 8)`.

Two binary trees are considered leaf-similar if their leaf value sequence is the same.

Return `true` if and only if the two given trees with head nodes `root1` and `root2` are leaf-similar.

Note:

• Both of the given trees will have between `1` and `100` nodes.

## 题目大意 #

• 给定的两颗树可能会有 1 到 200 个结点。
• 给定的两颗树上的值介于 0 到 200 之间。

## 解题思路 #

• 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。
• 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。

## 代码 #

``````func leafSimilar(root1 *TreeNode, root2 *TreeNode) bool {
leaf1, leaf2 := []int{}, []int{}
dfsLeaf(root1, &leaf1)
dfsLeaf(root2, &leaf2)
if len(leaf1) != len(leaf2) {
return false
}
for i := range leaf1 {
if leaf1[i] != leaf2[i] {
return false
}
}
return true
}

func dfsLeaf(root *TreeNode, leaf *[]int) {
if root != nil {
if root.Left == nil && root.Right == nil {
*leaf = append(*leaf, root.Val)
}
dfsLeaf(root.Left, leaf)
dfsLeaf(root.Right, leaf)
}
}
``````

Apr 8, 2023