Question

1602. 找到二叉树中最近的右侧节点

English Version

题目描述

给定一棵二叉树的根节点 root 和树中的一个节点 u ，返回与 u 所在层中距离最近的右侧节点，当 u 是所在层中最右侧的节点，返回 null 。

 

示例 1：

输入：root = [1,2,3,null,4,5,6], u = 4
输出：5
解释：节点 4 所在层中，最近的右侧节点是节点 5。

示例 2：

输入：root = [3,null,4,2], u = 2
输出：null
解释：2 的右侧没有节点。

示例 3：

输入：root = [1], u = 1
输出：null

示例 4：

输入：root = [3,4,2,null,null,null,1], u = 4
输出：2

 

提示:

• 树中节点个数的范围是 [1, 105] 。
• 1 <= Node.val <= 105
• 树中所有节点的值是唯一的。
• u 是以 root 为根的二叉树的一个节点。

解法

方法一：BFS

BFS 层序遍历，找到 $u$ 所在层的右侧相邻节点。

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

# 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 findNearestRightNode(self, root: TreeNode, u: TreeNode) -> Optional[TreeNode]:
    q = deque([root])
    while q:
      for i in range(len(q) - 1, -1, -1):
        root = q.popleft()
        if root == u:
          return q[0] if i else None
        if root.left:
          q.append(root.left)
        if root.right:
          q.append(root.right)

/**
 * 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 TreeNode findNearestRightNode(TreeNode root, TreeNode u) {
    Deque<TreeNode> q = new ArrayDeque<>();
    q.offer(root);
    while (!q.isEmpty()) {
      for (int i = q.size(); i > 0; --i) {
        root = q.pollFirst();
        if (root == u) {
          return i > 1 ? q.peekFirst() : null;
        }
        if (root.left != null) {
          q.offer(root.left);
        }
        if (root.right != null) {
          q.offer(root.right);
        }
      }
    }
    return null;
  }
}

/**
 * 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:
  TreeNode* findNearestRightNode(TreeNode* root, TreeNode* u) {
    queue<TreeNode*> q{{root}};
    while (q.size()) {
      for (int i = q.size(); i; --i) {
        root = q.front();
        q.pop();
        if (root == u) return i > 1 ? q.front() : nullptr;
        if (root->left) q.push(root->left);
        if (root->right) q.push(root->right);
      }
    }
    return nullptr;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func findNearestRightNode(root *TreeNode, u *TreeNode) *TreeNode {
  q := []*TreeNode{root}
  for len(q) > 0 {
    for i := len(q); i > 0; i-- {
      root = q[0]
      q = q[1:]
      if root == u {
        if i > 1 {
          return q[0]
        }
        return nil
      }
      if root.Left != nil {
        q = append(q, root.Left)
      }
      if root.Right != nil {
        q = append(q, root.Right)
      }
    }
  }
  return nil
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} u
 * @return {TreeNode}
 */
var findNearestRightNode = function (root, u) {
  const q = [root];
  while (q.length) {
    for (let i = q.length; i; --i) {
      root = q.shift();
      if (root == u) {
        return i > 1 ? q[0] : null;
      }
      if (root.left) {
        q.push(root.left);
      }
      if (root.right) {
        q.push(root.right);
      }
    }
  }
  return null;
};

方法二：DFS

DFS 先序遍历二叉树，首次搜索到 $u$ 时，标记目前层数 $d$，下次遇到同一层的节点时，即为目标节点。

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

# 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 findNearestRightNode(self, root: TreeNode, u: TreeNode) -> Optional[TreeNode]:
    def dfs(root, i):
      nonlocal d, ans
      if root is None or ans:
        return
      if d == i:
        ans = root
        return
      if root == u:
        d = i
        return
      dfs(root.left, i + 1)
      dfs(root.right, i + 1)

    d = 0
    ans = None
    dfs(root, 1)
    return ans

/**
 * 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 TreeNode u;
  private TreeNode ans;
  private int d;

  public TreeNode findNearestRightNode(TreeNode root, TreeNode u) {
    this.u = u;
    dfs(root, 1);
    return ans;
  }

  private void dfs(TreeNode root, int i) {
    if (root == null || ans != null) {
      return;
    }
    if (d == i) {
      ans = root;
      return;
    }
    if (root == u) {
      d = i;
      return;
    }
    dfs(root.left, i + 1);
    dfs(root.right, i + 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:
  TreeNode* u;
  TreeNode* ans;
  int d = 0;

  TreeNode* findNearestRightNode(TreeNode* root, TreeNode* u) {
    this->u = u;
    dfs(root, 1);
    return ans;
  }

  void dfs(TreeNode* root, int i) {
    if (!root || ans) return;
    if (d == i) {
      ans = root;
      return;
    }
    if (root == u) {
      d = i;
      return;
    }
    dfs(root->left, i + 1);
    dfs(root->right, i + 1);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func findNearestRightNode(root *TreeNode, u *TreeNode) *TreeNode {
  d := 0
  var ans *TreeNode
  var dfs func(*TreeNode, int)
  dfs = func(root *TreeNode, i int) {
    if root == nil || ans != nil {
      return
    }
    if d == i {
      ans = root
      return
    }
    if root == u {
      d = i
      return
    }
    dfs(root.Left, i+1)
    dfs(root.Right, i+1)
  }
  dfs(root, 1)
  return ans
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} u
 * @return {TreeNode}
 */
var findNearestRightNode = function (root, u) {
  let d = 0;
  let ans = null;
  function dfs(root, i) {
    if (!root || ans) {
      return;
    }
    if (d == i) {
      ans = root;
      return;
    }
    if (root == u) {
      d = i;
      return;
    }
    dfs(root.left, i + 1);
    dfs(root.right, i + 1);
  }
  dfs(root, 1);
  return ans;
};