05 August 2008
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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
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_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
Example 1:
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Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.
Example 2:
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Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself
according to the LCA definition.
Note:
求二叉搜索树的最小共同祖先,正常的思路是用递归来求解,由于二叉搜索树的特点是左<=根<=右,所以根节点的值一直都是中间值,大于左子树的所有节点值,小于右子树的所有节点值,所以如果根节点的值大于给定的两个值p和q之间的较大值,说明p和q都在左子树中,那么此时我们就进入根节点的左子节点继续递归寻找共同父节点;如果根节点小于p和q之间的较小值,说明p和q都在右子树中,那么就进入根节点的右子节点继续递归,如果都不是(大于p和q中较小值而小于较大值),则说明当前根节点就是最小共同父节点,直接返回;
如果是非递归的写法,就是把递归的过程用while来代替,但是每次循环需要更新一下当前的根节点。
递归
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) {
return null;
}
if (root.val < Math.min(p.val, q.val)) {//p,q都在右子树中
return lowestCommonAncestor(root.right, p, q);
} else if (root.val > Math.max(p.val, q.val)) {//p,q都在左子树中
return lowestCommonAncestor(root.left, p, q);
} else {//p,q一个在左一个在右,直接返回
return root;
}
}
}
非递归
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) {
return null;
}
while (true) {
if (root.val < Math.min(p.val, q.val)) {
root = root.right;
} else if (root.val > Math.max(p.val, q.val)) {
root = root.left;
} else {
break;
}
}
return root;
}
}