235. LCA of a binary search tree

Algorithm Comments

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).”

Example 1:

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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:

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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.
Given binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]

#236. LCA of a binary tree

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

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).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

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Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

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Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
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### Analyzation:
Use the theory of `divide and conquer`, for a node, if `p` and `q` exists in its left and right subtree respectively, it means such node will be `LCA`, otherwise, `p` or `q` could be `LCA`. 

Then we need to do traversal on the tree, if we meet `p` or `q`, we simply return it or we return null, if we found one node with no null returned from its left child and right child, then we find the `LCA`.

Time complexity will be `O(logn)` where `n` is the number of nodes. 

### Solution

/**

  • Definition for a binary tree node.

  • public class TreeNode {

  • int val;

  • TreeNode left;

  • TreeNode right;

  • TreeNode(int x) { val = x; }

  • }
    */
    class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    if(root == null || root == p || root == q){
    return root;
    }

     TreeNode left = this.lowestCommonAncestor(root.left, p, q);
 TreeNode right = this.lowestCommonAncestor(root.right, p, q);

 if(left != null && right != null){
     return root;
 }
 return left == null ? right : left;

    }
    }

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