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1457. Pseudo-Palindromic Paths in a Binary Tree

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Description

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

Solutions

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pseudoPalindromicPaths(self, root: TreeNode) -> int:
        def dfs(root):
            if root is None:
                return
            nonlocal ans, counter
            counter[root.val] += 1
            if root.left is None and root.right is None:
                if sum(1 for i in range(1, 10) if counter[i] % 2 == 1) < 2:
                    ans += 1
            else:
                dfs(root.left)
                dfs(root.right)
            counter[root.val] -= 1

        ans = 0
        counter = [0] * 10
        dfs(root)
        return ans

Java

/**
 * 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 {
    private int ans;
    private int[] counter;

    public int pseudoPalindromicPaths(TreeNode root) {
        ans = 0;
        counter = new int[10];
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        ++counter[root.val];
        if (root.left == null && root.right == null) {
            if (check(counter)) {
                ++ans;
            }
        } else {
            dfs(root.left);
            dfs(root.right);
        }
        --counter[root.val];
    }

    private boolean check(int[] counter) {
        int n = 0;
        for (int i = 1; i < 10; ++i) {
            if (counter[i] % 2 == 1) {
                ++n;
            }
        }
        return n < 2;
    }
}

C++

/**
 * 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 ans;
    vector<int> counter;

    int pseudoPalindromicPaths(TreeNode* root) {
        ans = 0;
        counter.resize(10);
        dfs(root);
        return ans;
    }

    void dfs(TreeNode* root) {
        if (!root) return;
        ++counter[root->val];
        if (!root->left && !root->right) {
            int n = 0;
            for (int i = 1; i < 10; ++i)
                if (counter[i] % 2 == 1)
                    ++n;
            if (n < 2) ++ans;
        } else {
            dfs(root->left);
            dfs(root->right);
        }
        --counter[root->val];
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pseudoPalindromicPaths(root *TreeNode) int {
	ans := 0
	counter := make([]int, 10)
	var dfs func(root *TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		counter[root.Val]++
		if root.Left == nil && root.Right == nil {
			n := 0
			for i := 1; i < 10; i++ {
				if counter[i]%2 == 1 {
					n++
				}
			}
			if n < 2 {
				ans++
			}
		} else {
			dfs(root.Left)
			dfs(root.Right)
		}
		counter[root.Val]--
	}
	dfs(root)
	return ans
}

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