Leaf-Similar Trees Problem
Description
LeetCode Problem 872.
Consider all the leaves of a binary tree, fromleft to right order, the values of thoseleaves 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-similarif 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:
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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:
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Input: root1 = [1], root2 = [1]
Output: true
Example 3:
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Input: root1 = [1], root2 = [2]
Output: false
Example 4:
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Input: root1 = [1,2], root2 = [2,2]
Output: true
Example 5:
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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].
Sample C++ Code
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/**
* 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> &leafs) {
if ((root->left == NULL) && (root->right == NULL))
leafs.push_back(root->val);
if (root->left != NULL)
dfs(root->left, leafs);
if (root->right != NULL)
dfs(root->right, leafs);
}
bool leafSimilar(TreeNode* root1, TreeNode* root2) {
vector<int> leafs1;
vector<int> leafs2;
if (root1 != NULL)
dfs(root1, leafs1);
if (root2 != NULL)
dfs(root2, leafs2);
return (leafs1 == leafs2);
}
};