1602. Find Nearest Right Node in Binary Tree

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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:

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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:

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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, 10^5].
  • 1 <= Node.val <= 10^5
  • All values in the tree are distinct .
  • u is a node in the binary tree rooted at root.

Hints/Notes

  • use targetDepth as a proxy to show the node has been found

Solution

Language: C++

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/**
 * 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:
    int targetDepth = -1;
    TreeNode* target = nullptr;

    TreeNode* findNearestRightNode(TreeNode* root, TreeNode* u) {
        traverse(root, u, 0);
        return target;
    }

    void traverse(TreeNode* root, TreeNode* u, int depth) {
        if (!root || target) {
            return;
        }
        if (depth == targetDepth) {
            target = root;
        }
        if (root == u) {
            targetDepth = depth;
        }
        traverse(root->left, u, depth + 1);
        traverse(root->right, u, depth + 1);
    }
};