题目
最近公共祖先
给定一棵二叉树,找到两个节点的最近公共父节点(LCA)。
最近公共祖先是两个节点的公共的祖先节点且具有最大深度。
样例
对于下面这棵二叉树
4
/ \
3 7
/ \
5 6
LCA(3, 5) = 4
LCA(5, 6) = 7
LCA(6, 7) = 7
解题
不知道如何下手,参考链接,自顶向下的解法中有下面的说明:
初首先看看3和5,这两个节点分居根节点4的两侧,如果可以从子节点往父 节点递推,那么他们将在根节点4处第一次重合;再来看看5和6,这两个都在根节点4的右侧,沿着父节点往上递推,他们将在节点7处第一次重合;最 后来看看6和7,此时由于7是6的父节点,故7即为所求。从这三个基本例子我们可以总结出两种思路——自顶向下(从前往后递推)和自底向上(从后 往前递推)。 顺着上述实例的分析,我们首先看看自底向上的思路,自底向上的实现用一句话来总结就是——如果遍历到的当前节点是 A/B 中的任意一个,那么我们就向父节点汇报此节点,否则递归到节点为空时返回空值。具体来说会有如下几种情况: 1.当前节点不是两个节点中的任意一个,此时应判断左右子树的返回结果。 1.若左右子树均返回非空节点,那么当前节点一定是所求的根节点,将当前节点逐层向前汇报。// 两个节点分居树的两侧 2.若左右子树仅有一个子树返回非空节点,则将此非空节点向父节点汇报。// 节点仅存在于树的一侧 3.若左右子树均返回NULL, 则向父节点返回NULL. // 节点不在这棵树中 2.当前节点即为两个节点中的一个,此时向父节点返回当前节点
Java
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: The root of the binary search tree. * @param A and B: two nodes in a Binary. * @return: Return the least common ancestor(LCA) of the two nodes. */ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) { // write your code here if( root ==null || A==root || B == root) return root; TreeNode left = lowestCommonAncestor(root.left,A,B); TreeNode right = lowestCommonAncestor(root.right,A,B); if(left != null && right !=null) return root; if(left!=null) return left; return right; } }
Python
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
""" Definition of TreeNode: class TreeNode: def __init__(self, val): self.val = val self.left, self.right = None, None """ import copy class Solution: """ @param root: The root of the binary search tree. @param A and B: two nodes in a Binary. @return: Return the least common ancestor(LCA) of the two nodes. """ def lowestCommonAncestor(self, root, A, B): # write your code here if root == None or root == A or root == B: return root left = self.lowestCommonAncestor(root.left,A,B) right = self.lowestCommonAncestor(root.right,A,B) if left!=None and right!=None: return root if left!=None: return left return right
这个博客给了dfs的解法,但是用java一直写不对,在LeetCode discuss 找到了根据DFS从根节点找到当前节点的路径,知道路径就很简单了,这里还是看程序吧
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: The root of the binary search tree. * @param A and B: two nodes in a Binary. * @return: Return the least common ancestor(LCA) of the two nodes. */ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) { // write your code here if( root ==null || A==root || B == root) return root; ArrayList<TreeNode> pathA = new ArrayList<TreeNode>(); ArrayList<TreeNode> pathB = new ArrayList<TreeNode>(); dfs(root,pathA,A); dfs(root,pathB,B); TreeNode res = null; for(int i=0;i<Math.min(pathA.size(),pathB.size());i++){ TreeNode pa = pathA.get(i); TreeNode pb = pathB.get(i); if(pa == pb){ res = pa; }else{ break; } } return res; } public boolean dfs(TreeNode root,ArrayList<TreeNode> path,TreeNode node){ if(root == null) return false; if(node == root){ // 最后一个节点也要加进去,不然会出错 path.add(root); return true; } path.add(root); if(dfs(root.left,path,node) ==true) return true; if(dfs(root.right,path,node) ==true) return true; path.remove(path.size() -1); return false; } }