1602. Find Nearest Right Node in Binary Tree 🔒

Description

Given the root of a binary tree and a node u in the tree, return the nearest node on the same level that is to the right of u, or return null if u is the rightmost node in its level.

Example 1:

Input: root = [1,2,3,null,4,5,6], u = 4
Output: 5
Explanation: The nearest node on the same level to the right of node 4 is node 5.

Example 2:

Input: root = [3,null,4,2], u = 2
Output: null
Explanation: There are no nodes to the right of 2.

Constraints:

The number of nodes in the tree is in the range [1, 10⁵].
1 <= Node.val <= 10⁵
All values in the tree are distinct.
u is a node in the binary tree rooted at root.

Solutions

Solution 1: BFS

We can use Breadth-First Search, starting from the root node. When we reach node $u$, we return the next node in the queue.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the binary tree.

Python3JavaC++GoJavaScript

# 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;
};

Solution 2: DFS

DFS performs a pre-order traversal of the binary tree. The first time we search to node $u$, we mark the current depth $d$. The next time we encounter a node at the same level, it is the target node.