448. Inorder Successor in Binary Search Tree [LintCode]

Given a binary search tree (See Definition) and a node in it, find the in-order successor of that node in the BST.

If the given node has no in-order successor in the tree, returnnull.

Notice

It's guaranteed_p_is one node in the given tree. (You can directly compare the memory address to find p)

Example

Given tree =[2,1]and node =1:

  2
 /
1

return node2.

Given tree =[2,1,3]and node =2:

  2
 / \
1   3

return node3.

第一个while循环是在利用BST的特性从原始节点（root of tree）去寻找p点所在的位置，基于中序遍历的特性，如果所找的p在左边，则用successor变量记录当前节点为左子树的后续节点。

找到p点之后，根据中序遍历的特性，去寻找下一个节点所在位置：如果p有右子树则successor为右子树的左下角元素；如果p没有右子树则successor为上面所记载的successor。

/**
 * 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 inorderSuccessor(TreeNode root, TreeNode p) {
        TreeNode successor = null;
        while(root != null && root.val != p.val) {
            if (p.val < root.val) {
                successor = root;
                root = root.left;
            }
            if (root.val < p.val) {
                root = root.right;
            }
        }
        if (root == null) {
            return null;
        }
        if (root.right == null) {
            return successor;
        }
        root = root.right;
        while (root.left != null) {
            root = root.left;
        }
        return root;
    }
}