Leetcode Lowest Common Ancestor of a 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 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.

---

解题思路：

首先要注意什么是Binary Search Tree(即左孩子比它小，右孩子比它大)，根据这个规律找到算法。

1.递归。我喜欢这个，简洁明了！！！记住。

(1) P, Q都比root小，则LCA在左树，我们继续在左树中寻找LCA

(2) P, Q都比root大，则LCA在右树，我们继续在右树中寻找LCA

(3) 其它情况，表示P,Q在root两边，或者二者其一是root，或者都是root，这些情况表示root就是LCA，直接返回root即可。

2. 非递归。

---

Java code

1. recursion

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        //recursion
        if(root == null || (root.val >= Math.min(p.val, q.val) && root.val <= Math.max(p.val,q.val))) {
            return root;
        }
        if(root.val > Math.max(p.val, q.val)) {
            return lowestCommonAncestor(root.left, p, q);
        }else {
            return lowestCommonAncestor(root.right, p, q);    
        }
    }
}

2. non recursion

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        TreeNode curr = root;
        TreeNode p_next, q_next;
        while(curr != null) {
            if(p.val > curr.val){
                p_next = curr.right;
            }else if(p.val == curr.val) {
                p_next = curr;
            }else{
                p_next = curr.left;
            }
            
            if(q.val > curr.val) {
                q_next = curr.right;
            }else if(q.val == curr.val) {
                q_next = curr;
            }else {
                q_next = curr.left;
            }
            
            if(p_next != q_next){
                return curr;
            }
            curr = p_next;
        }
        return curr;
    }
}

---

Reference:

1. https://leetcode.com/discuss/52241/3-lines-java-concise-and-easy-understanding-solution

2. https://leetcode.com/discuss/50981/accepted-non-recursive-c-code