Leetcode 1457. Pseudo-Palindromic Paths in a Binary Tree (二叉树遍历题)
- Pseudo-Palindromic Paths in a Binary Tree
Medium
Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be pseudo-palindromic if at least one permutation of the node values in the path is a palindrome.
Return the number of pseudo-palindromic paths going from the root node to leaf nodes.
Example 1:
Input: root = [2,3,1,3,1,null,1]
Output: 2
Explanation: The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: the red path [2,3,3], the green path [2,1,1], and the path [2,3,1]. Among these paths only red path and green path are pseudo-palindromic paths since the red path [2,3,3] can be rearranged in [3,2,3] (palindrome) and the green path [2,1,1] can be rearranged in [1,2,1] (palindrome).
Example 2:
Input: root = [2,1,1,1,3,null,null,null,null,null,1]
Output: 1
Explanation: The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: the green path [2,1,1], the path [2,1,3,1], and the path [2,1]. Among these paths only the green path is pseudo-palindromic since [2,1,1] can be rearranged in [1,2,1] (palindrome).
Example 3:
Input: root = [9]
Output: 1
Constraints:
The number of nodes in the tree is in the range [1, 105].
1 <= Node.val <= 9
解法1：
注意：这里的path是指从root到leaf。
怎么才能算pseudo-palindromic呢? 只要统计出1-9每个数字的个数，如果奇数个的个数<=1就算！
/**
* 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:
int pseudoPalindromicPaths (TreeNode* root) {
pathNum.resize(10, 0);
helper(root);
return count;
}
private:
vector<int> pathNum;
int count = 0;
void helper(TreeNode *root) {
if (!root) return;
pathNum[root->val]++;
if (!root->left && !root->right) {
int oddNum = 0, evenNum = 0;
for (int i = 1; i <= 9; i++) {
if (pathNum[i] & 0x1) oddNum++;
else evenNum++;
}
if (oddNum <= 1) {
count++;
//return;
}
pathNum[root->val]--; //记得这里也要--
return;
}
helper(root->left);
helper(root->right);
pathNum[root->val]--;
return;
}
};
解法2：思路跟上面差不多，但是用XOR。
注意:
- bitwise的操作优先级都很低，要加括号。比如
(pathXor & (pathXor - 1)) == 0
pathXor ^= (0x1 << (root->val)); - 要用 pathXor ^= (0x1 << (root->val))， 不能直接用pathXor ^= root->val。
否则如果有多个数的个数是奇数的话，pathXor看不出来。
用pathXor ^= (0x1 << (root->val))的话，1-9每个数字个数是奇数还是偶数就一目了然了。 - int x。 x & (x-1)会抹掉最后一个1。那么如果x=0的话，是不是也成立呢? 是的，因为0&(-1)=0.
所以，如果pathXor有一个1，或者为全0，那么pathXor & (pathXor - 1)都是0。这个就可以作为有pseudo-palindromic的评判标准。
/**
* 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:
int pseudoPalindromicPaths (TreeNode* root) {
helper(root);
return count;
}
private:
int pathXor = 0;
int count = 0;
void helper(TreeNode *root) {
if (!root) return;
pathXor ^= (0x1 << (root->val));
if (!root->left && !root->right) {
if ((pathXor & (pathXor - 1)) == 0) count++;
pathXor ^= (0x1 << (root->val));
return;
}
helper(root->left);
helper(root->right);
pathXor ^= (0x1 << (root->val));
return;
}
};