Lowest Common Ancestor of a Binary Search Tree

update Jun 27, 2017

leetcode

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.

思路

基本的想法是要利用到其中BST的性质，即最近的祖先的值一定是在两个target nodes之间。

Java Code：

    // java
    public class Solution {
        TreeNode ret = null;
        public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
            if (root == null) return null;
            if (root.val < p.val && root.val < q.val)
                return lowestCommonAncestor(root.right, p, q);
            if (root.val > p.val && root.val > q.val)
                return lowestCommonAncestor(root.left, p, q);
            else
                return root;
        }  
    }

update 2018-06-29 16:08:42

Iterative Java Solution

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (p.val > q.val) {
            TreeNode t = p;
            p = q;
            q = t;
        }
        TreeNode node = root;
        while (! (p.val <= node.val && q.val >= node.val)) {
            if (node.val < p.val) node = node.right;
            else node = node.left;
        }
        return node;
    }
}

update Jul 3 2021

更新一个 Iterative 做法

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        int lower = Math.min(p.val, q.val);
        int upper = Math.max(p.val, q.val);
        while (true) {
            if (root.val < lower) root = root.right;
            else if (root.val > upper) root = root.left;
            else return root;
        }
    }
}