---
tags: leetcode
---

# [1740. Find Distance in a Binary Tree](https://leetcode.com/problems/find-distance-in-a-binary-tree/)
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# My Solution
## The Key Idea for Solving This Coding Question
## C++ Code

/**
 * 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 findDistance(TreeNode *root, int p, int q) {
        if (p == q) {
            return 0;
        }

        TreeNode *lca = findLCA(root, p, q);

        int pLen = 0;
        if (root->val != p) {
            pLen = bfs(lca, p);
        }

        int qLen = 0;
        if (root->val != q) {
            qLen  = bfs(lca, q);
        }

        return pLen + qLen;
    }

private:
    TreeNode *findLCA(TreeNode *root, int p, int q) {
        if (root == nullptr) {
            return nullptr;
        }

        if (root->val == p || root->val == q) {
            return root;
        }

        TreeNode *leftSubtree = findLCA(root->left, p, q);
        TreeNode *rightSubtree = findLCA(root->right, p, q);

        if (leftSubtree == nullptr) {
            return rightSubtree;
        }
        if (rightSubtree == nullptr) {
            return leftSubtree;
        }
        return root;
    }

    int bfs(TreeNode *root, int val) {
        if (root == nullptr) {
            return 0;
        }

        int len = 0;
        queue<TreeNode *> q;

        q.push(root);
        while (!q.empty()) {
            int qLen = q.size();

            for (int i = 0; i < qLen; ++i) {
                TreeNode *curr = q.front();
                q.pop();

                if (curr->val == val) {
                    return len;
                }

                if (curr->left != nullptr) {
                    q.push(curr->left);
                }
                if (curr->right != nullptr) {
                    q.push(curr->right);
                }
            }
            ++len;
        }

        return -1;
    }
};

## Time Complexity
$O(n)$
$n$ is the number of nodes in the binary tree.
## Space Complexity
$O(n)$
$n$ is the number of nodes in the binary tree.
# Miscellane
<!--
# Test Cases

[3,5,1,6,2,0,8,null,null,7,4]
5
0

[3,5,1,6,2,0,8,null,null,7,4]
5
7

[3,5,1,6,2,0,8,null,null,7,4]
5
5

-->