[LeetCode 235] Lowest Common Ancestor of a Binary Search Tree

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本文介绍了如何在一个二叉搜索树中找到两个给定节点的最近公共祖先。通过利用二叉搜索树的性质，提供了一个有效的解决方案，并包括了验证节点是否为有效树节点的步骤。

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.

Solution:

1use property of BST

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

Add check to verify if p and q is valid tree node

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null) return null;
        int max = root.val;
        int min = root.val;
        TreeNode t1 = root;
        while(t1!=null){
            min = t1.val;
            t1 = t1.left;
        }
        t1 = root;
        while(t1!=null){
            max = t1.val;
            t1 = t1.right;
        }
        if(p.val>q.val && (min>q.val || max<p.val)) return null;
        if(q.val>p.val && (min>p.val || max<q.val)) return null;
        if((root.val>=p.val && root.val<=q.val)||(root.val<=p.val && root.val>=q.val))
            return root;
        if(root.val>=p.val && root.val>=q.val){
            return lowestCommonAncestor(root.left, p, q);
        }
        return lowestCommonAncestor(root.right, p, q);
    }