1457. Pseudo-Palindromic Paths in a Binary Tree (M)

Pseudo-Palindromic Paths in a Binary Tree (M)

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:

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 given binary tree will have between 1 and 10^5 nodes.
• Node values are digits from 1 to 9.

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题意

在二叉树中找到一条从根结点到叶结点的路径，使路径上的数字能按照某种顺序组成回文序列，求这样的路径的个数。

思路

回溯法。到达叶结点时，判断当前路径上出现次数为奇数的数字是否只有0个或1个，是的话说明能构成回文序列。

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代码实现

Java

class Solution {
    public int pseudoPalindromicPaths (TreeNode root) {
        return dfs(root, 0, new boolean[10]);
    }

    private int dfs(TreeNode root, int oddNum, boolean[] isOdd) {
        oddNum += isOdd[root.val] ? -1 : 1;
        
        if (root.left == null && root.right == null) {
            return oddNum <= 1 ? 1 : 0;
        }

        isOdd[root.val] = !isOdd[root.val];   // 注意在这里改变状态，如果放在最上面，在叶结点判断返回之前要恢复原状态
        int leftCount = root.left != null ? dfs(root.left, oddNum, isOdd) : 0;
        int rightCount = root.right != null ? dfs(root.right, oddNum, isOdd) : 0;
        isOdd[root.val] = !isOdd[root.val];

        return leftCount + rightCount;
    }
}