Inorder Successor in BST

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

Note: If the given node has no in-order successor in the tree, return 

null

.

Example 1:

Input: root =          [2,1,3], p = 1

  2
 / \
1   3
                
Output: 2      

Example 2:

Input: root =          [5,3,6,2,4,null,null,1], p = 6

      5
     / \
    3   6
   / \
  2   4
 /   
1
                
Output: null      

思路: 首先找到P点，同时记录过程中最后一个左拐的点，也就是有可能是大于p的第一个点 （如果p没有右边儿子的话）

如果p有右边儿子，那么就是右边儿子最左边的那个node，否则就是last left turn node；

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        if(root == null || p == null) {
            return null;
        }
        TreeNode pre = null;
        TreeNode node = root;
        while(node != p) {
            if(node.val > p.val) {
                pre = node;
                node = node.left;
            } else if(node.val < p.val) {
                node = node.right;
            } else {
                break;
            }
        }
        if(node.right != null) {
            return findLeft(node.right);
        } else {
            return pre;
        }
    }
    
    private TreeNode findLeft(TreeNode node) {
        TreeNode res = null;
        while(node != null) {
            res = node;
            node = node.left;
        }
        return res;
    }
}