[LintCode] 915. Inorder Predecessor in BST

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

Example

Example1

Input: root = {2,1,3}, p = 1
Output: null

Example2

Input: root = {2,1}, p = 2
Output: 1

Notice

If the given node has no in-order predecessor in the tree, return null

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: the given BST
     * @param p: the given node
     * @return: the in-order predecessor of the given node in the BST
     */
    public TreeNode inorderPredecessor(TreeNode root, TreeNode p) {
        // write your code here
        TreeNode lowestLeftP = null;
        TreeNode cur = root;
        if (root == null) {
            return null;
        }
        while (cur != p) {
            if (cur.val < p.val) {
                lowestLeftP = cur;
                cur = cur.right;
            } else {
                cur = cur.left;
            }
        }
        TreeNode son = cur.left;
        TreeNode res = son;
        while (son != null) {
            res = son;
            son = son.right;
        }
        if (res != null) {
            return res;
        }
        return lowestLeftP;
    }
}