面试题之树中两个节点的最低公共祖先节点

1.树为二叉搜索树，二叉搜索树特点：值：右>根>左

①思路

从根节点开始遍历

a：若该节点值比所给两个节点均大，则公共祖先节点必在其左子树上，遍历其左子树。

b：若该节点值比所给两个节点均小，则公共祖先节点必在其右子树上，遍历其右子树。

c：直到找到一个节点，位于两节点的中间。

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)."

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.

②代码

public class BanarySearchTreeMain {public static void main(String[] args) {// TODO Auto-generated method stubTreeNode root = initTree();TreeNode result = lowestCommonAncestor(root, new TreeNode(2), new TreeNode(4));System.out.println(result.val);}/* * 初始化二叉搜索树，其特点：右>根>左； */private static TreeNode initTree(){TreeNode root = new TreeNode(6);TreeNode node0 = new TreeNode(0);TreeNode node2 = new TreeNode(2);TreeNode node3 = new TreeNode(3);TreeNode node4 = new TreeNode(4);TreeNode node5 = new TreeNode(5);TreeNode node7 = new TreeNode(7);TreeNode node8 = new TreeNode(8);TreeNode node9 = new TreeNode(9);root.left = node2;root.right = node8;node2.left = node0;node2.right = node4;node8.left = node7;node8.right = node9;node4.left = node3;node4.right = node5;return root;}/* * 二叉搜索树最低公共父节点 */private static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if(root==null||root==p||root==q) {return root;}if(root.val>p.val&&root.val>q.val) {return lowestCommonAncestor(root.left, p, q);}if(root.val<p.val&&root.val<q.val) {return lowestCommonAncestor(root.right, p, q);}else {return root;}}}class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;}}

2.树为普通二叉树

①思路

a：遍历左子树，若与两节点中某值相等或左右均不为null，则返回根节点，否则，继续遍历；

b：遍历右子树，若与两节点中某值相等或左右均不为null，则返回根节点，否则，继续遍历；

c：返回节点；

For example,the lowest common ancestor (LCA) of nodes 5 and 1 is 3.Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

②代码

public class CommonTreeMain {public static void main(String[] args) {TreeNode root = initTree();TreeNode result = lowestCommonAncestor(root, new TreeNode(5), new TreeNode(4));System.out.println(result.val);}private static TreeNode initTree(){TreeNode root = new TreeNode(3);TreeNode node0 = new TreeNode(0);TreeNode node1 = new TreeNode(1);TreeNode node2 = new TreeNode(2);TreeNode node4 = new TreeNode(4);TreeNode node5 = new TreeNode(5);TreeNode node6 = new TreeNode(6);TreeNode node7 = new TreeNode(7);TreeNode node8 = new TreeNode(8);root.left = node5;root.right = node1;node5.left = node6; node5.right = node2; node1.left = node0;node1.right = node8;node2.left = node7;node2.right = node4;return root;}/* * 查找最低公共祖先节点 */    private static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {          if(root==null||root.val==p.val||root.val==q.val)              return root;          TreeNode left = lowestCommonAncestor(root.left,p,q);          TreeNode right = lowestCommonAncestor(root.right,p,q);          if (left == null)               return right;          if (right == null)               return left;          return root;      }  }

③迭代算法：

需要我们保存下由root根节点到p和q节点的路径，并且将路径存入list中，则问题转化为求两个list集合的最后一个共同元素。

    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;            }  

转载：http://blog.csdn.net/qq_25827845/article/details/74612786