题目描述

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

题目链接

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

方法思路

Approach1:

class Solution {
    //Runtime: 4 ms, faster than 100.00%
    //Memory Usage: 35.1 MB, less than 34.41% 
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(find(p, q.val)) return p;
        if(find(q, p.val)) return q;
        while(root != null){
            if((root.val > p.val && root.val < q.val) || 
               (root.val < p.val && root.val > q.val))
                break;
            if(root.val > p.val && root.val > q.val){
                root = root.left;
                continue;
            }
            if(root.val < p.val && root.val < q.val)
                root = root.right;
        }
        return root;
        
    }
    public boolean find(TreeNode root, int val){
        if(root == null) return false;
        while(root != null){
            if(root.val == val) return true;
            if(root.val > val) 
                root = root.left;
            else
                root = root.right;
        }
        return false;
    }
}

Approach2:recursive

class Solution {
    //Runtime: 4 ms, faster than 100.00% 
    //Memory Usage: 35.8 MB, less than 5.11%
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {

        // Value of current node or parent node.
        int parentVal = root.val;

        // Value of p
        int pVal = p.val;

        // Value of q;
        int qVal = q.val;

        if (pVal > parentVal && qVal > parentVal) {
            // If both p and q are greater than parent
            return lowestCommonAncestor(root.right, p, q);
        } else if (pVal < parentVal && qVal < parentVal) {
            // If both p and q are lesser than parent
            return lowestCommonAncestor(root.left, p, q);
        } else {
            // We have found the split point, i.e. the LCA node.
            return root;
        }
    }
}

Approach3: iterative, 思路和方法二一样

class Solution {
    //Runtime: 4 ms, faster than 100.00% 
    //Memory Usage: 35.2 MB, less than 27.21% 
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {

        // Value of p
        int pVal = p.val;

        // Value of q;
        int qVal = q.val;

        // Start from the root node of the tree
        TreeNode node = root;

        // Traverse the tree
        while (node != null) {

            // Value of ancestor/parent node.
            int parentVal = node.val;

            if (pVal > parentVal && qVal > parentVal) {
                // If both p and q are greater than parent
                node = node.right;
            } else if (pVal < parentVal && qVal < parentVal) {
                // If both p and q are lesser than parent
                node = node.left;
            } else {
                // We have found the split point, i.e. the LCA node.
                return node;
            }
        }
        return null;
    }
}