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993. Cousins in Binary Tree

Description

Given the root of a binary tree with unique values and the values of two different nodes of the tree x and y, return true if the nodes corresponding to the values x and y in the tree are cousins, or false otherwise.

Two nodes of a binary tree are cousins if they have the same depth with different parents.

Note that in a binary tree, the root node is at the depth 0, and children of each depth k node are at the depth k + 1.

 

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

 

Constraints:

• The number of nodes in the tree is in the range [2, 100].
• 1 <= Node.val <= 100
• Each node has a unique value.
• x != y
• x and y are exist in the tree.

Solutions

BFS or DFS.

• Java
• C++
• Python
• Go
• TypeScript

• /**
   * 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 x, y;
      private TreeNode p1, p2;
      private int d1, d2;
  
      public boolean isCousins(TreeNode root, int x, int y) {
          this.x = x;
          this.y = y;
          dfs(root, null, 0);
          return p1 != p2 && d1 == d2;
      }
  
      private void dfs(TreeNode root, TreeNode p, int d) {
          if (root == null) {
              return;
          }
          if (root.val == x) {
              p1 = p;
              d1 = d;
          }
          if (root.val == y) {
              p2 = p;
              d2 = d;
          }
          dfs(root.left, root, d + 1);
          dfs(root.right, root, d + 1);
      }
  }
  

• /**
   * 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:
      bool isCousins(TreeNode* root, int x, int y) {
          TreeNode *p1, *p2;
          int d1, d2;
          function<void(TreeNode*, TreeNode*, int)> dfs = [&](TreeNode* root, TreeNode* fa, int d) {
              if (!root) {
                  return;
              }
              if (root->val == x) {
                  p1 = fa;
                  d1 = d;
              }
              if (root->val == y) {
                  p2 = fa;
                  d2 = d;
              }
              dfs(root->left, root, d + 1);
              dfs(root->right, root, d + 1);
          };
          dfs(root, nullptr, 0);
          return p1 != p2 && d1 == d2;
      }
  };
  

• # 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 isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
          def dfs(root, fa, d):
              if root is None:
                  return
              if root.val == x:
                  t[0] = (fa, d)
              if root.val == y:
                  t[1] = (fa, d)
              dfs(root.left, root, d + 1)
              dfs(root.right, root, d + 1)
  
          t = [None, None]
          dfs(root, None, 0)
          return t[0][0] != t[1][0] and t[0][1] == t[1][1]
  
  

• /**
   * Definition for a binary tree node.
   * type TreeNode struct {
   *     Val int
   *     Left *TreeNode
   *     Right *TreeNode
   * }
   */
  func isCousins(root *TreeNode, x int, y int) bool {
  	var p1, p2 *TreeNode
  	var d1, d2 int
  	var dfs func(*TreeNode, *TreeNode, int)
  	dfs = func(root *TreeNode, fa *TreeNode, d int) {
  		if root == nil {
  			return
  		}
  		if root.Val == x {
  			p1, d1 = fa, d
  		}
  		if root.Val == y {
  			p2, d2 = fa, d
  		}
  		dfs(root.Left, root, d+1)
  		dfs(root.Right, root, d+1)
  	}
  	dfs(root, nil, 0)
  	return p1 != p2 && d1 == d2
  }
  

• /**
   * Definition for a binary tree node.
   * class TreeNode {
   *     val: number
   *     left: TreeNode | null
   *     right: TreeNode | null
   *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
   *         this.val = (val===undefined ? 0 : val)
   *         this.left = (left===undefined ? null : left)
   *         this.right = (right===undefined ? null : right)
   *     }
   * }
   */
  
  function isCousins(root: TreeNode | null, x: number, y: number): boolean {
      let [d1, d2] = [0, 0];
      let [p1, p2] = [null, null];
      const q: [TreeNode, TreeNode][] = [[root, null]];
      for (let depth = 0; q.length > 0; ++depth) {
          const t: [TreeNode, TreeNode][] = [];
          for (const [node, parent] of q) {
              if (node.val === x) {
                  [d1, p1] = [depth, parent];
              } else if (node.val === y) {
                  [d2, p2] = [depth, parent];
              }
              if (node.left) {
                  t.push([node.left, node]);
              }
              if (node.right) {
                  t.push([node.right, node]);
              }
          }
          q.splice(0, q.length, ...t);
      }
      return d1 === d2 && p1 !== p2;
  }
  
  

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