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;
        }
    }
    
  • // OJ: https://leetcode.com/problems/find-distance-in-a-binary-tree/
    // Time: O(N)
    // Space: O(H)
    class Solution {
        bool findPath(TreeNode *root, int target, vector<TreeNode*> &path) {
            if (!root) return false;
            path.push_back(root);
            if (root->val == target || findPath(root->left, target, path) || findPath(root->right, target, path)) return true;
            path.pop_back();
            return false;
        }
    public:
        int findDistance(TreeNode* root, int p, int q) {
            if (p == q) return 0;
            vector<TreeNode*> a, b;
            findPath(root, p, a);
            findPath(root, q, b);
            int i = 0;
            while (i < a.size() && i < b.size() && a[i] == b[i]) ++i;
            return a.size() + b.size() - 2 * i;
        }
    };
    
  • print("Todo!")
    

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