GuilinDev
Lc1457
05 August 2008
1457 Pseudo-Palindromic Paths in a Binary Tree
题目
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:
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2
3
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:
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2
3
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:
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2
Input: root = [9]
Output: 1
Constraints:
- The given binary tree will have between
1
1
1
10^5
- Node values are digits from
1
1
1
9
分析
Top Down DFS
代码
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
int result;
public int pseudoPalindromicPaths (TreeNode root) {
result = 0;
if (root == null) {
return result;
}
int[] path = new int[10];
helper(root, path);
return result;
}
private void helper(TreeNode node, int[] path) {
if (node == null) {
return;
}
if (node.left == null && node.right == null) {
path[node.val]++;
if (checkPP(path)) {
result++;
}
path[node.val]--;
return;
}
path[node.val]++;
helper(node.left, path);
helper(node.right, path);
path[node.val]--;
}
private boolean checkPP(int[] path) {
int countOdd = 0;
for (int i = 1; i < path.length; i++) {
if (path[i] % 2 != 0) {
countOdd++;
}
if (countOdd >= 2) {
return false;
}
}
return true;
}
}