【Leetcode】Lowest Common Ancestor of a Binary Search Tree
2015-07-20 21:05
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【题目】
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
Another example is LCA of nodes
since a node can be a descendant of itself according to the LCA definition.
【思路】
开始一看题。。我去。。懵了,感觉这个easy的题目猛然并么有什么思路。看了discuss里面大神们的解答又顿时恍然大悟了。
首先注意是BST!!说明什么。。sorted. 左子树全是比根小得,右子树全是比根大的。
先判断,p,q是不是同一颗树上的,如果不是同一个树上的,说明一个比根小,一个比根大,那他们不会有比root更low的相同点了,就是root了!
如果是同一个树上的,比如两者都比root小,那么他们的更low的root就在左子树上。如果都比根大,那么就
【代码】
return ((p.val-root.val)*(q.val-root.val)<=0) ? root : lowestCommonAncestor(p.val>root.val?root.right:root.left, p, q);
正常版本。。。
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
2and
8is
6.
Another example is LCA of nodes
2and
4is
2,
since a node can be a descendant of itself according to the LCA definition.
【思路】
开始一看题。。我去。。懵了,感觉这个easy的题目猛然并么有什么思路。看了discuss里面大神们的解答又顿时恍然大悟了。
首先注意是BST!!说明什么。。sorted. 左子树全是比根小得,右子树全是比根大的。
先判断,p,q是不是同一颗树上的,如果不是同一个树上的,说明一个比根小,一个比根大,那他们不会有比root更low的相同点了,就是root了!
如果是同一个树上的,比如两者都比root小,那么他们的更low的root就在左子树上。如果都比根大,那么就
【代码】
return ((p.val-root.val)*(q.val-root.val)<=0) ? root : lowestCommonAncestor(p.val>root.val?root.right:root.left, p, q);
正常版本。。。
public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if(root==null) return null; if(p.val<root.val&&q.val<root.val) { return lowestCommonAncestor(root.left,p,q); } if(p.val>root.val&&q.val>root.val) { return lowestCommonAncestor(root.right,p,q); }else{ return root; } } }
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