Formatted question description: https://leetcode.ca/all/1740.html

1740. Find Distance in a Binary Tree

Level

Medium

Description

Given the root of a binary tree and two integers p and q, return the distance between the nodes of value p and value q in the tree.

The distance between two nodes is the number of edges on the path from one to the other.

Example 1:

Image text

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 0

Output: 3

Explanation: There are 3 edges between 5 and 0: 5-3-1-0.

Example 2:

Image text

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 7

Output: 2

Explanation: There are 2 edges between 5 and 7: 5-2-7.

Example 3:

Image text

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 5

Output: 0

Explanation: The distance between a node and itself is 0.

Constraints:

  • The number of nodes in the tree is in the range [1, 10^4].
  • 0 <= Node.val <= 10^9
  • All Node.val are unique.
  • p and q are values in the tree.

Solution

Related: 236-Lowest-Common-Ancestor-of-a-Binary-Tree/

First, find the lowest common ancestor of the two nodes with values p and q. Next, calculate the distances from the lowest common ancestor to the two nodes with values p and q, respectively. Finally, calculate the sum of the two distances and return.

Alternative-1

hashmap to store distance of every node-pair during finding LCA, then just map look up.

Alternative-2

Dist(n1, n2) = Dist(root, n1) + Dist(root, n2) - 2 * Dist(root, lca)

  • ‘n1’ and ‘n2’ are the two given keys
  • ‘root’ is root of given Binary Tree.
  • ‘lca’ is lowest common ancestor of n1 and n2
  • Dist(n1, n2) is the distance between n1 and n2.
/**
 * 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;
 *     }
 * }
 */

// alternative: hashmap to store distance of every node-pair during finding LCA, then just map look up

class Solution {
    public int findDistance(TreeNode root, int p, int q) {
        if (p == q)
            return 0;
        TreeNode ancestor = lowestCommonAncestor(root, p, q);
        return getDistance(ancestor, p) + getDistance(ancestor, q);
    }

    public TreeNode lowestCommonAncestor(TreeNode root, int p, int q) {
        if (root == null)
            return null;
        if (root.val == p || root.val == q)
            return root;
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if (left != null && right != null)
            return root;
        return left == null ? right : left;
    }

    public int getDistance(TreeNode node, int val) {
        if (node == null)
            return -1;
        if (node.val == val)
            return 0;
        int leftDistance = getDistance(node.left, val);
        int rightDistance = getDistance(node.right, val);
        int subDistance = Math.max(leftDistance, rightDistance);
        return subDistance >= 0 ? subDistance + 1 : -1;
    }
}

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