LeetCode //C - 235. Lowest Common Ancestor of a Binary Search Tree

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235. Lowest Common Ancestor of a Binary Search Tree

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

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

Example 1:

Input: root = $6,2,8,0,4,7,9,null,null,3,5$ , p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = $6,2,8,0,4,7,9,null,null,3,5$ , p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = $2,1$ , p = 2, q = 1
Output: 2

Constraints:

The number of nodes in the tree is in the range $2 , 1 0 5$ $2, 10\^5$ $2,105$ .
− 1 0 9 < = N o d e . v a l < = 1 0 9 -10^9 <= Node.val <= 10^9 −109<=Node.val<=109
All Node.val are unique.
p != q
p and q will exist in the BST.

From: LeetCode

Link: 233. Number of Digit One

Solution:

Ideas:

Start from the root and keep traversing the tree.
Depending on the values of p and q relative to the current node, move left or right.
If you find a node where p and q lie on different sides (or one of them is equal to the current node), that node is the LCA.

Code:

c 复制代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
    while (root != NULL) {
        // If both p and q are greater than root, LCA lies in the right subtree
        if (p->val > root->val && q->val > root->val) {
            root = root->right;
        }
        // If both p and q are smaller than root, LCA lies in the left subtree
        else if (p->val < root->val && q->val < root->val) {
            root = root->left;
        }
        // We have found the split point, i.e. the LCA node
        else {
            return root;
        }
    }
    return NULL; // this line will never be reached if p and q are guaranteed to be in the tree
}