lintcode448- Inorder Successor in Binary Search Tree- medium

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, return null.

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 node 2.

Given tree = [2,1,3] and node = 2:

  2
 / 
1   3

return node 3.

Challenge 

O(h), where h is the height of the BST.

1. 递归法。

a. 如果root.val大了，答案要么去左边找，要么是当前root点(p是左子树的右下角最后一个点，这种情况课巧妙地用返回null来标记)。想清楚a非常重要！

b.如果root.val等于了，那要去右子树找

c.如果root.val小了，那要去右子树找(p都在右边，p的后继更在右边了)

2. 也递归法，处理方法更intuitive但麻烦：

要查找的点是否有右孩子：如果有，简单，直接找右子树的最小节点。如果没有，则找到比该节点大且相差最小的父节点（可用stack回溯）。

1. 九章实现的简洁算法，学

// version: 高频题班
public class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        // write your code here
        if (root == null || p == null) {
            return null;
        }

        if (root.val <= p.val) {
            return inorderSuccessor(root.right, p);
        } else {
            TreeNode left = inorderSuccessor(root.left, p);
            return (left != null) ? left : root;
        }
    }
}

2.自己实现的stack回溯。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */


public class Solution {
    /*
     * @param root: The root of the BST.
     * @param p: You need find the successor node of p.
     * @return: Successor of p.
     */
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        // write your code here
        return helper(root, p, new Stack<TreeNode>());
    }
    
    public TreeNode helper(TreeNode root, TreeNode p, Stack<TreeNode> stack) {
        if (root == null) {
            return null;
        }
        if (p.val > root.val) {
            stack.push(root);
            return helper(root.right, p, stack);
        } else if (p.val < root.val) {
            stack.push(root);
            return helper(root.left, p, stack);
        } else {
            if (p.right != null) {
                TreeNode successor = p.right;
                while (successor.left != null) {
                    successor = successor.left;
                }
                return successor;
            } else {
                if (stack.isEmpty()) {
                    return null;
                }
                TreeNode successor = stack.pop();
                while (successor.val <= p.val && !stack.isEmpty()) {
                    successor = stack.pop();
                }
                // 处理当p是最后一个点的情况。
                if (successor.val <= p.val) {
                    successor = null;
                }
                return successor;
            }
        }
    }
}