LeetCode #872 LeafSimilarTree. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Leaf-Similar Trees.
Memory Usage: 14.3 MB, less than 31.80% of C++ online submissions for Leaf-Similar Trees.
LeetCode #872
Q.
Consider all the leaves of a binary tree. From left to right order, the values of those leaves form a leaf value sequence.
이진트리의 모든 마지막 노드들을 고려한다. 왼쪽부터 오른쪽까지, 그러한 끄트머리 노드들 값은 leaf 값 배열을 구성한다.
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.
예를들면, 위에보면, leaf 값들 배열은 (6, 7, 4, 9, 8) 이다.
두개의 이진트리들은 각각의 leaf 값 배열이 같으면 leaf-similar 하다고 여겨진다.
두개의 주어진 트리들이 leaf-similar 하면 true, 아니면 false 를 반환해라.
Note:
- Both of the given trees will have between 1 and 100 nodes.
노트 :
주어지는 트리들의 노드 개수는 1~ 100개이다.
Process
// Process
//1. Input two rootNodes
//2. Make each leafSequence using recursive
//3. Check if two leafSeq are the same or not, and get result
//4. Return result
// 처리과정
//1. 두개의 루트노드를 입력받는다.
//2. 각각의 끄트머리 노드 배열 ( leaf sequence ) 를 만든다. (재귀)
//3. 각각의 leaf seq 가 같은지 확인해서 결과값을 구한다.
//4. 결과를 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool leafSimilar(TreeNode* root1, TreeNode* root2) {
bool answer = true;
vector<int> leafSeq1;
vector<int> leafSeq2;
recursiveMakeLeaf(leafSeq1, root1);
recursiveMakeLeaf(leafSeq2, root2);
int length = 0;
if (leafSeq1.size() > leafSeq2.size())
length = leafSeq2.size();
else
length = leafSeq1.size();
for (int i = 0; answer && i < length; ++i) {
if (leafSeq1[i] != leafSeq2[i])
answer = false;
}
return answer;
}
private:
void recursiveMakeLeaf(vector<int>& leafSeq, TreeNode* node) {
if (node != 0)
if (node->left == 0 && node->right == 0) {
leafSeq.push_back(node->val);
} else {
recursiveMakeLeaf(leafSeq, node->left);
recursiveMakeLeaf(leafSeq, node->right);
}
}
};
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