java_lca_of_binary_search_tree

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 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).”

Examples

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,$10^5$].
• $-10^9$ <= Node.val <= $10^9$
• All Node.val are unique.
• p != q
• p and q will exist in the BST.

解析

題目給定一個二元搜尋樹的根結點，還有兩個結點p, q

要求 p, q 兩個結點的最小共同祖先結點

要解出這題需要先了解二元搜尋樹的特性

二元搜尋樹的特性如下

1. 每個結點的值比左子樹下每個結點的值大
2. 每個結點的值比右子樹下每個結點的值小

所以這題要找最小共同祖先

這個共同祖先 root 必須符合幾個特性

1 Min(p.Val, q.Val) ≤ root.Val

2 Max(p.Val, q.Val) ≥ root.Val

因為是二元搜尋樹所以再搜詢時

可以照以下條件搜尋

當 Min(p.Val, q.Val) > root.Val ，向右子樹搜訊

當 Max(p.Val, q.Val) < root.Val ，向左子樹搜訊

如果都不再兩個條件代表當下結點就是共同組結點，因為由根結點往下搜尋，因此會是最小子結點

程式碼

class Solution {
  /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode(int x) { val = x; }
   * }
   */
  public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    TreeNode small = p, large = q;
    if (p.val > q.val) {
      small = q;
      large = p;
    }
    if (root.val > large.val) {
      return lowestCommonAncestor(root.left, small, large);
    }
    if (root.val < small.val) {
      return lowestCommonAncestor(root.right, small, large);
    }
    return root;
  }
}

困難點

1. 需要理解 BST(二元搜尋樹)特性
2. 需要理解 definition of LCA on Wikipedia 最小共同祖結點定義

Solve Point

• Understand what problem need to be solved
• Analysis Complexity