[LeetCode] 1457. Pseudo-Palindromic Paths in a Binary Tree

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.

二叉树中的伪回文路径。

给你一棵二叉树，每个节点的值为 1 到 9 。我们称二叉树中的一条路径是 「伪回文」的，当它满足：路径经过的所有节点值的排列中，存在一个回文序列。

请你返回从根到叶子节点的所有路径中 伪回文 路径的数目。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/pseudo-palindromic-paths-in-a-binary-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是DFS + 回溯。这道题跟113题很像，你需要去找到树中每一条从根节点到叶子节点的路径，然后再多一步判断，判断每一条路径是不是满足题目定义的伪回文。判断伪回文额外写一个函数判断即可。这里我提供两种做法，思路差不多，第一种方法速度快一些，第二种方法不使用全局变量。两种做法的时间空间复杂度均为O(n)。

DFS + 回溯

count是全局变量而且hashmap是被带入helper函数往下一层去递归的，所以需要回溯

Java实现

/**
 * 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 count = 0;

    public int pseudoPalindromicPaths(TreeNode root) {
        int[] map = new int[10];
        helper(root, map);
        return count;
    }

    private void helper(TreeNode root, int[] map) {
        // base case
        if (root == null) {
            return;
        }
        map[root.val]++;
        if (root.left == null && root.right == null) {
            if (isPalindrome(map)) {
                count++;
            }
        }

        helper(root.left, map);
        helper(root.right, map);
        map[root.val]--;
    }

    private boolean isPalindrome(int[] map) {
        int single = 0;
        for (int i = 0; i < map.length; i++) {
            if (map[i] % 2 == 1) {
                single++;
            }
        }
        return single > 1 ? false : true;
    }
}

更像前序遍历的一种做法，没有回溯

Java实现

/**
 * 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 {
    public int pseudoPalindromicPaths(TreeNode root) {
        return helper(root, new HashSet<>());
    }

    private int helper(TreeNode root, HashSet<Integer> set) {
        if (root == null) {
            return 0;
        }
        if (set.contains(root.val)) {
            set.remove(root.val);
        } else {
            set.add(root.val);
        }
        if (root.left == null && root.right == null) {
            return set.size() <= 1 ? 1 : 0;
        }
        // 注意这里hashset的细节
        int left = helper(root.left, new HashSet<>(set));
        int right = helper(root.right, new HashSet<>(set));
        return left + right;
    }
}

相关题目

113. Path Sum II

1457. Pseudo-Palindromic Paths in a Binary Tree

LeetCode 题目总结