# 20200115

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).”

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

6
/ \
2 8
/\ /\
0 4 7 9
/\
3 5

// Binary Search Tree特性，root會大於等於左邊，root會小於右邊。
// pq大於左右邊，祖先必在左
// pq小於左右邊，祖先必在右
// 反之回傳自己
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root.val > p.val && root.val > q.val){
            return lowestCommonAncestor(root.left, p, q);
        } else if(root.val < p.val && root.val < q.val){
            return lowestCommonAncestor(root.right, p, q);
        } else{
            return root;
        }
    }

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]
p = 5, q = 4 output: 5

3
/ \
5 1
/\ /\
6 2 0 8
/\
7 4

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
 // 找到就先回自己，找左傳右找右傳左回自己
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || root == p || root == q) {
            return root;
        }
        
        TreeNode left = lowestCommonAncestor(root.left, p, q); 
        TreeNode right = lowestCommonAncestor(root.right, p, q); 
        if (left == null) {
            return right;
        }
        
        if (right == null) {
            return left;
        }
        
        return root;
    }
}