49 lines
1.7 KiB
Java
49 lines
1.7 KiB
Java
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# 68. 树中两个节点的最低公共祖先
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## 解题思路
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### 二叉查找树
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[Leetcode : 235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/)
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二叉查找树中，两个节点 p, q 的公共祖先 root 满足 root.val >= p.val && root.val <= q.val。
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<div align="center"> <img src="pics/047faac4-a368-4565-8331-2b66253080d3.jpg" width="220"/> </div><br>
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```java
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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if (root == null)
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return root;
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if (root.val > p.val && root.val > q.val)
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return lowestCommonAncestor(root.left, p, q);
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if (root.val < p.val && root.val < q.val)
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return lowestCommonAncestor(root.right, p, q);
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return root;
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}
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```
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### 普通二叉树
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[Leetcode : 236. Lowest Common Ancestor of a Binary Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/)
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在左右子树中查找是否存在 p 或者 q，如果 p 和 q 分别在两个子树中，那么就说明根节点就是最低公共祖先。
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<div align="center"> <img src="pics/d27c99f0-7881-4f2d-9675-c75cbdee3acd.jpg" width="250"/> </div><br>
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```java
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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if (root == null || root == p || root == q)
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return root;
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TreeNode left = lowestCommonAncestor(root.left, p, q);
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TreeNode right = lowestCommonAncestor(root.right, p, q);
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return left == null ? right : right == null ? left : root;
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}
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```
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<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-1.png"></img></div>
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