LeetCode题解：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 v and w as the lowest node in T that has both v and w as descendants (where we allow a 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.

题意：给定一个二叉树，找到两个节点NA， NB的最近公共祖先(LCA)。
比如对于下图，4 和 7 的 LCA 是6， 1和13的LCA 是 8。

解决思路：首先二叉搜索树要求了，左子树所有节点的值均小于根结点的值（非空），右子树反之。那么，我们可以得到的是，如果p,q均在root的左子树或右子树上，p.val-root.val与q.val-root.val的值符号是相同的，只有分列左右时才会异号。

代码：

递归

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        return ((p.val-root.val)*(q.val-root.val)<=0) ? root : lowestCommonAncestor(p.val>root.val?root.right:root.left, p, q);
    }

非递归

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        while (true) {
            if (root.val > p.val && root.val > q.val)
                root = root.left;
            else if (root.val < p.val && root.val < q.val)
                root = root.right;
            else
                return root;
        }
    }