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  • Lowest Common Ancestor of a Binary Search Tree

    1. Title

    235. Lowest Common Ancestor of a Binary Search Tree

    2. Http address

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

    3. The question

    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.

     4 My code(AC)

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        
        public boolean DFS(TreeNode root, TreeNode target, LinkedList<TreeNode> path)
        {
        
            if (root == null || target == null)
                return false;
    
            path.addLast(root);
            if (root == target) {
                return true;
            }
    
            if (DFS(root.left, target, path)) {
                return true;
            } else {
                if (root.left != null) {
                    path.removeLast();
                }
                boolean re =  DFS(root.right, target, path);
                if( root.right != null && re == false){
                    path.removeLast();
                }
                return re;    
            }
        
        }
        
        public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
            
            
            if( p == null || q == null)
            {
                return null;
            }
            LinkedList<TreeNode> path1 = new LinkedList<TreeNode>();
            LinkedList<TreeNode> path2 = new LinkedList<TreeNode>();
            DFS(root,p,path1);
            DFS(root,q,path2);
            
            if( path1.size() <= 0 || path2.size() <= 0)
                return null;
            TreeNode bf = path1.removeFirst();
            path2.removeFirst();
            while( !path1.isEmpty() && !path2.isEmpty()){
                TreeNode node1 = path1.removeFirst();
                TreeNode node2 = path2.removeFirst();
                if( node1 != node2){
                    break;
                }
                bf =  node1;
            }
            return bf;
        
        }
    }
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  • 原文地址:https://www.cnblogs.com/ordili/p/5515744.html
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