872. Leaf-Similar Trees

Andreea
2 min readMay 15, 2021

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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.

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], root2 = [1]
Output: true

Example 3:

Input: root1 = [1], root2 = [2]
Output: false

Example 4:

Input: root1 = [1,2], root2 = [2,2]
Output: true

Example 5:

Input: root1 = [1,2,3], root2 = [1,3,2]
Output: false
/**
* 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 leaf(TreeNode *root, vector<int>& v){

if(root->right == NULL && root->left == NULL){

v.push_back(root->val);
}

if(root->right != NULL){
leaf(root->right, v);
}
if(root->left != NULL){
leaf(root->left, v);
}

}

bool leafSimilar(TreeNode* root1, TreeNode* root2) {

vector<int> v1;
leaf(root1, v1);
vector<int> v2;
leaf(root2, v2);

if(v1.size() != v2.size()){

return false;
}
for(int i=0; i<v1.size(); i++){

if(v1[i] != v2[i]){

return false;
}
}
return true;
}
};

Runtime: 4 ms, faster than 68.51% of C++ online submissions for Leaf-Similar Trees.

Memory Usage: 12.9 MB, less than 20.96% of C++ online submissions for Leaf-Similar Trees.

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