Question

993. 二叉树的堂兄弟节点

English Version

题目描述

在二叉树中，根节点位于深度 0 处，每个深度为 k 的节点的子节点位于深度 k+1 处。

如果二叉树的两个节点深度相同，但 父节点不同 ，则它们是一对_堂兄弟节点_。

我们给出了具有唯一值的二叉树的根节点 root ，以及树中两个不同节点的值 x 和 y 。

只有与值 x 和 y 对应的节点是堂兄弟节点时，才返回 true 。否则，返回 false。

 

示例 1：

输入：root = [1,2,3,4], x = 4, y = 3
输出：false

示例 2：

输入：root = [1,2,3,null,4,null,5], x = 5, y = 4
输出：true

示例 3：

输入：root = [1,2,3,null,4], x = 2, y = 3
输出：false

 

提示：

• 二叉树的节点数介于 2 到 100 之间。
• 每个节点的值都是唯一的、范围为 1 到 100 的整数。

 

解法

方法一：BFS

我们定义一个队列 $q$，队列中存储的是节点和其父节点。初始时，将根节点和空节点放入队列中。

每次从队列中取出一个节点，如果该节点的值为 $x$ 或 $y$，则记录该节点的父节点和深度。如果该节点的左右子节点不为空，则将左右子节点和该节点放入队列中。

当队列中所有节点都处理完毕后，如果 $x$ 和 $y$ 的深度相同且父节点不同，则返回 $true$，否则返回 $false$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

# 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:
    q = deque([(root, None)])
    depth = 0
    p1 = p2 = None
    d1 = d2 = None
    while q:
      for _ in range(len(q)):
        node, parent = q.popleft()
        if node.val == x:
          p1, d1 = parent, depth
        elif node.val == y:
          p2, d2 = parent, depth
        if node.left:
          q.append((node.left, node))
        if node.right:
          q.append((node.right, node))
      depth += 1
    return p1 != p2 and d1 == d2

/**
 * 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 boolean isCousins(TreeNode root, int x, int y) {
    Deque<TreeNode[]> q = new ArrayDeque<>();
    q.offer(new TreeNode[] {root, null});
    int d1 = 0, d2 = 0;
    TreeNode p1 = null, p2 = null;
    for (int depth = 0; !q.isEmpty(); ++depth) {
      for (int n = q.size(); n > 0; --n) {
        TreeNode[] t = q.poll();
        TreeNode node = t[0], parent = t[1];
        if (node.val == x) {
          d1 = depth;
          p1 = parent;
        } else if (node.val == y) {
          d2 = depth;
          p2 = parent;
        }
        if (node.left != null) {
          q.offer(new TreeNode[] {node.left, node});
        }
        if (node.right != null) {
          q.offer(new TreeNode[] {node.right, node});
        }
      }
    }
    return p1 != p2 && d1 == d2;
  }
}

/**
 * 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) {
    queue<pair<TreeNode*, TreeNode*>> q;
    q.push({root, nullptr});
    int d1 = 0, d2 = 0;
    TreeNode *p1 = nullptr, *p2 = nullptr;
    for (int depth = 0; q.size(); ++depth) {
      for (int n = q.size(); n; --n) {
        auto [node, parent] = q.front();
        q.pop();
        if (node->val == x) {
          d1 = depth;
          p1 = parent;
        } else if (node->val == y) {
          d2 = depth;
          p2 = parent;
        }
        if (node->left) {
          q.push({node->left, node});
        }
        if (node->right) {
          q.push({node->right, node});
        }
      }
    }
    return d1 == d2 && p1 != p2;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func isCousins(root *TreeNode, x int, y int) bool {
  type pair struct{ node, parent *TreeNode }
  var d1, d2 int
  var p1, p2 *TreeNode
  q := []pair{{root, nil}}
  for depth := 0; len(q) > 0; depth++ {
    for n := len(q); n > 0; n-- {
      node, parent := q[0].node, q[0].parent
      q = q[1:]
      if node.Val == x {
        d1, p1 = depth, parent
      } else if node.Val == y {
        d2, p2 = depth, parent
      }
      if node.Left != nil {
        q = append(q, pair{node.Left, node})
      }
      if node.Right != nil {
        q = append(q, pair{node.Right, node})
      }
    }
  }
  return d1 == d2 && p1 != p2
}

/**
 * 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;
}

方法二：DFS

我们设计一个函数 $dfs(root, parent, depth)$，表示从根节点 $root$ 出发，其父节点为 $parent$，深度为 $depth$，进行深度优先搜索。

在函数中，我们首先判断当前节点是否为空，如果为空，则直接返回。如果当前节点的值为 $x$ 或 $y$，则记录该节点的父节点和深度。然后对当前节点的左右子节点分别调用函数 $dfs$，其中父节点为当前节点，深度为当前深度加 $1$。即 $dfs(root.left, root, depth + 1)$ 和 $dfs(root.right, root, depth + 1)$。

当整棵二叉树遍历完毕后，如果 $x$ 和 $y$ 的深度相同且父节点不同，则返回 $true$，否则返回 $false$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

# 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, parent, depth):
      if root is None:
        return
      if root.val == x:
        st[0] = (parent, depth)
      elif root.val == y:
        st[1] = (parent, depth)
      dfs(root.left, root, depth + 1)
      dfs(root.right, root, depth + 1)

    st = [None, None]
    dfs(root, None, 0)
    return st[0][0] != st[1][0] and st[0][1] == st[1][1]

/**
 * 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 int d1, d2;
  private TreeNode p1, p2;

  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 parent, int depth) {
    if (root == null) {
      return;
    }
    if (root.val == x) {
      d1 = depth;
      p1 = parent;
    } else if (root.val == y) {
      d2 = depth;
      p2 = parent;
    }
    dfs(root.left, root, depth + 1);
    dfs(root.right, root, depth + 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) {
    int d1, d2;
    TreeNode* p1;
    TreeNode* p2;
    function<void(TreeNode*, TreeNode*, int)> dfs = [&](TreeNode* root, TreeNode* parent, int depth) {
      if (!root) {
        return;
      }
      if (root->val == x) {
        d1 = depth;
        p1 = parent;
      } else if (root->val == y) {
        d2 = depth;
        p2 = parent;
      }
      dfs(root->left, root, depth + 1);
      dfs(root->right, root, depth + 1);
    };
    dfs(root, nullptr, 0);
    return p1 != p2 && d1 == d2;
  }
};

/**
 * 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 d1, d2 int
  var p1, p2 *TreeNode
  var dfs func(root, parent *TreeNode, depth int)
  dfs = func(root, parent *TreeNode, depth int) {
    if root == nil {
      return
    }
    if root.Val == x {
      d1, p1 = depth, parent
    } else if root.Val == y {
      d2, p2 = depth, parent
    }
    dfs(root.Left, root, depth+1)
    dfs(root.Right, root, depth+1)
  }
  dfs(root, nil, 0)
  return d1 == d2 && p1 != p2
}

/**
 * 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, p1, p2] = [0, 0, null, null];
  const dfs = (root: TreeNode | null, parent: TreeNode | null, depth: number) => {
    if (!root) {
      return;
    }
    if (root.val === x) {
      [d1, p1] = [depth, parent];
    } else if (root.val === y) {
      [d2, p2] = [depth, parent];
    }
    dfs(root.left, root, depth + 1);
    dfs(root.right, root, depth + 1);
  };
  dfs(root, null, 0);
  return d1 === d2 && p1 !== p2;
}