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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    Example 1:

    1. Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
    2. Output: 6
    3. Explanation: The LCA of nodes 2 and 8 is 6.

    Example 2:

    1. Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
    2. Output: 2
    3. 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.

    题意

    在二叉查找树BST中找到两个指定结点的最近共同祖先。

    思路

    根据定义,两个结点分列于共同祖先的左右两侧(也可以共同祖先重合),又有BST的排列规则,可以有以下搜索方法:如果p、q的值都小于当前结点,说明共同祖先只可能在左子树;如果p、q的值都大于当前结点,说明共同祖先只可能在右子树;如果p值小于当前结点、q值大于当前结点,或p值大于当前结点、q值小于当前结点,或p、q中一点就是当前结点,说明当前结点就是要找的共同祖先。


    代码实现

    1. class Solution {
    2. public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    3. while (root != null) {
    4. if (p.val < root.val && q.val < root.val) {
    5. root = root.left;
    6. } else if (p.val > root.val && q.val > root.val) {
    7. root = root.right;
    8. } else {
    9. break;
    10. }
    11. }
    12. return root;
    13. }
    14. }