leetcode 235: Lowest Common Ancestor of a Binary Search Tree
Lowest Common Ancestor of a Binary Search Tree
Total Accepted: 203 Total Submissions: 511Given 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]
如果如果p,q 比root小, 则LCA必定在左子树, 如果p,q比root大, 则LCA必定在右子树. 如果一大一小, 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;
}
}这道题还可以有一个followup. 如果是普通二叉树, 而不是BST. 则应该遍历节点, 先找到p,q. 同时记录下从root到该几点的路径. 之后比较路径,最后一个相同的节点便是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 pathp = new ArrayList<>();
List 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 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;
}
}