1457. Pseudo-Palindromic Paths in a Binary Tree.md

1457. Pseudo-Palindromic Paths in a Binary Tree

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leetcode Daily Challenge on December 29th, 2020.

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Difficulty : Medium

Related Topics : DFS、Bit Manipulation、Tree

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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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Solution

• mine
  • Java
    • DFS & HashSet Runtime: 12 ms, faster than 70.36%, Memory Usage: 60.3 MB, less than 100.00% of Java online submissions

      //O(N)time O(Depth)space
      public int res = 0;
      public int pseudoPalindromicPaths (TreeNode root) {
          HashSet<Integer> set = new HashSet<>();
          dfs(root,set);
          return res;
      }
      
      public void dfs(TreeNode node, HashSet<Integer> set){
          if(node == null){
              return;
          }
          boolean in = set.contains(node.val);
          if(in){
              set.remove(node.val);
          }else{
              set.add(node.val);
          }
          if(node.left == null && node.right == null){
              res += set.size() > 1 ? 0 : 1;
          }else{
              dfs(node.left, set);
              dfs(node.right, set);
          }
          if(in){
              set.add(node.val);
          }else{
              set.remove(node.val);
          }
      }
      

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• the most votes
  • Bit Manipulation Runtime: 2 ms, faster than 100.00%, Memory Usage: 57.2 MB, less than 100.00% of Java online submissions

    //O(N)time O(1)space
    public int pseudoPalindromicPaths (TreeNode root) {
        return dfs(root, 0);
    }
    
    private int dfs(TreeNode root, int count) {
        if (root == null) return 0;
        count ^= 1 << (root.val - 1);
        int res = dfs(root.left, count) + dfs(root.right, count);
        if (root.left == root.right && (count & (count - 1)) == 0) res++;
        return res;
    }
    

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