Last Updated on 2022-12-08 by Clay
題目
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.
Example 1:
Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8] Output: true
Example 2:
Input: root1 = [1,2,3], root2 = [1,3,2] Output: false
Constraints:
- The number of nodes in each tree will be in the range
[1, 200]. - Both of the given trees will have values in the range
[0, 200].
題目會給定兩顆結構不同的樹，我們要判斷的是其葉子（亦即不帶有子節點的節點）從左到右的序列是否一致。如果兩棵樹的葉子序列相同，返回 true；不同，則返回 false。
解題思路
DFS
使用 DFS 走訪的話，便會理所當然地按照順序從左到右拜訪所有葉子節點。
我曾考慮過是否有一種先判斷第一棵樹 DFS 走到第一個葉子節點、接著再判斷第二顆樹 DFS 走到的一個葉子節點兩者是否相同，接著則是第一棵樹繼續往下走訪到第二個葉子節點… 依此類推。
但是考慮了一會兒似乎有點難寫，不確定是否有大神使用單線程寫出來了，照理來說應該可以… 但我還是首先採用暴力解了。
最直接的解法就是 DFS 走完兩顆樹，並在每一個節點時判斷底下是否有其他節點，若無，則使用一陣列將其保存起來；最後，則直接判斷兩棵樹的葉子節點陣列是否一致即可。
DFS 複雜度
| Time Complexity | O(n) |
| Space Complexity | O(n) |
C++ 範例程式碼
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void DFS(TreeNode* root, vector<int>& leaves) {
// Base case
if (!root) {
return;
}
// If the root is a leaf
if (!root->left && !root->right) {
leaves.emplace_back(root->val);
}
// DFS
DFS(root->left, leaves);
DFS(root->right, leaves);
}
bool leafSimilar(TreeNode* root1, TreeNode* root2) {
vector<int> leaves1;
vector<int> leaves2;
DFS(root1, leaves1);
DFS(root2, leaves2);
return leaves1 == leaves2;
}
};Python 範例程式碼
# 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 DFS(self, root: Optional[TreeNode], leaves: List[int]) -> None:
# Base case
if not root:
return
# If root is a leaf
if not root.left and not root.right:
leaves.append(root.val)
self.DFS(root.left, leaves)
self.DFS(root.right, leaves)
def leafSimilar(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
v1 = []
v2 = []
self.DFS(root1, v1)
self.DFS(root2, v2)
return v1 == v2