Lowest Common Ancestor of a Binary Search Tree #LeetCode

Problem Reference:
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

Problem Statement:

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Solution:

Binary tree traversal.

Termination case :
1). find LCA in left subtree
2). find LCA in right subtree
3). find LCA at current Node, (left + current, right + current, left + right)

Declare a struct Result to log searching result
struct Result {
  bool isPFind;
  bool isQFind;
}
or use a integer where bit is used to indicate whether node is found or not.

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { result = NULL; find(root, p, q); return result; } unsigned find(TreeNode* root, TreeNode* p, TreeNode* q) { if (!root) { return 0; } unsigned left = find(root->left, p, q); unsigned current = 0 | (root == p ? 1 : 0 ) | (root == q ? 2 : 0); unsigned right = find(root->right, p, q); if (left == 3) { return 3; } else if (right == 3) { return 3; } else if ((left | current | right) == 3) { result = root; return 3; } return left | current | right; } TreeNode* result; };