leetcode 235: Lowest Common Ancestor of a Binary Search Tree

Lowest Common Ancestor of a Binary Search Tree

Total Accepted: 203 Total Submissions: 511

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 allowa 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]

假设假设p,q 比root小, 则LCA必然在左子树, 假设p,q比root大, 则LCA必然在右子树. 假设一大一小, 则root即为LCA.

[CODE]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    //2, 1
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null || p==null || q==null) return null;
        
        if(Math.max(p.val, q.val) < root.val) {
            return lowestCommonAncestor(root.left, p, q);
        } else if(Math.min(p.val, q.val) > root.val) {
            return lowestCommonAncestor(root.right, p, q);
        } else return root;
    }
}

这道题还能够有一个followup. 假设是普通二叉树, 而不是BST.  则应该遍历节点, 先找到p,q. 同一时候记录下从root到该几点的路径.   之后比較路径,最后一个同样的节点便是LCA.

[CODE]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    //2, 1
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null || p==null || q==null) return null;
        
        List<TreeNode> pathp = new ArrayList<>();
        List<TreeNode> pathq = new ArrayList<>();
        pathp.add(root);
        pathq.add(root);
        
        getPath(root, p, pathp);
        getPath(root, q, pathq);
        
        TreeNode lca = null;
        for(int i=0; i<pathp.size() && i<pathq.size(); i++) {
            if(pathp.get(i) == pathq.get(i)) lca = pathp.get(i);
            else break;
        }
        return lca;
    }
    
    private boolean getPath(TreeNode root, TreeNode n, List<TreeNode> path) {
        if(root==n) {
            return true;
        }
        
        if(root.left!=null) {
            path.add(root.left);
            if(getPath(root.left, n, path)) return true;
            path.remove(path.size()-1);
        }
        
        if(root.right!=null) {
            path.add(root.right);
            if(getPath(root.right, n, path)) return true;
            path.remove(path.size()-1);
        }
        
        return false;
    }
}