# LeetCode.235.Lowest Common Ancestor of BST BST的最近公共祖先

## 题目描述

235 Lowest Common Ancestor of a Binary Search Tree
https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

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]

BST

Example 1:

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:

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:

• All of the nodes’ values will be unique.
• p and q are different and both values will exist in the BST.

## 解题过程

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (null == root) {
return root;
}
if (containNodes(root.left, p, q)) {
return lowestCommonAncestor(root.left, p, q);
} else if(containNodes(root.right, p, q)) {
return lowestCommonAncestor(root.right, p, q);
} else {
return root;
}
}

// 非递归先序遍历，判断root中是否包含p和q
public boolean containNodes(TreeNode root, TreeNode p, TreeNode q) {
if (null == root) {
return false;
}
boolean hasP = false;
boolean hasQ = false;
stack.push(root);
TreeNode cur = root;
while (!stack.isEmpty() || null != cur) {
while (null != cur) {
hasP = cur == p || hasP;
hasQ = cur == q || hasQ;
cur = cur.left;
if (null != cur) {
stack.push(cur);
}
}
cur = stack.pop();
if (null != cur.right) {
stack.push(cur.right);
}
cur = cur.right;
}
return hasP && hasQ;
}

## GitHub代码

algorithms/leetcode/leetcode/_235_LowestCommonAncestorOfBST.java
https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_235_LowestCommonAncestorOfBST.java

585

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