leetcode235. Lowest Common Ancestor of a Binary Search Tree

235. 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.

解法一

如果比root小，查找root左端；如果比root大，查找root右端。

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {        if (root == null) {            return root;        }        if (root.val < p.val && root.val < q.val) {            root = root.right;        } else if (root.val > p.val && root.val > q.val) {            root = root.left;        }        if (root == null) {            return root;        }        TreeNode left = lowestCommonAncestor(root.left, p, q);        TreeNode right = lowestCommonAncestor(root.right, p, q);        if (root == p || root == q) {            return root;        }        if (left != null && right !=null) {            return root;        }        if (left != null) {            return left;        }        if (right != null) {            return right;        }        return null;     }}

解法二

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {        TreeNode result = root;        if (root.val < p.val && root.val < q.val) {            result = lowestCommonAncestor(root.right, p, q);        } else if (root.val > p.val && root.val > q.val) {            result = lowestCommonAncestor(root.left, p, q);        } else {            result = root;        }        return result;    }}

解法三

针对解法一的改进，将相等提前到前面，如果一左一右直接返回root

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {        if (root == null) {            return root;        }        if (root.val < p.val && root.val < q.val) {            root = root.right;        } else if (root.val > p.val && root.val > q.val) {            root = root.left;        } else {            return root;        }        if (root == null) {            return root;        }        if (root == p || root == q) {            return root;        }        TreeNode left = lowestCommonAncestor(root.left, p, q);        TreeNode right = lowestCommonAncestor(root.right, p, q);        if (left != null && right !=null) {            return root;        }        if (left != null) {            return left;        }        if (right != null) {            return right;        }        return null;     }}