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文章目录
- 一、题目
- 二、题解
一、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.
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].
二、题解
/*** 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 preOrder(TreeNode* root,vector<int>& leaf){if(!root) return;if(root->left == nullptr && root->right == nullptr) leaf.push_back(root->val);preOrder(root->left,leaf);preOrder(root->right,leaf);}bool leafSimilar(TreeNode* root1, TreeNode* root2) {vector<int> leaf1;vector<int> leaf2;preOrder(root1,leaf1);preOrder(root2,leaf2);if(leaf1.size() != leaf2.size()) return false;for(int i = 0;i < leaf1.size();i++){if(leaf1[i] != leaf2[i]) return false;}return true;}
};
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