Medium 285题 Inorder Successor in BST

Question:

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

Solution:

分两种情况讨论，参考了https://segmentfault.com/a/1190000003792039，存下所有路径

/**
 * 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) {
        Stack st=new Stack();
        while(p!=root){
            st.push(root);
            if(root.val>p.val)
                root=root.left;
            else
                root=root.right;
        }
        if(p.right!=null)
            return FindMostLeft(p.right);
        else{
            return Find(p,st);
        }
            
    }
    public TreeNode FindMostLeft(TreeNode node)
    {
        while(node.left!=null)
            node=node.left;
        return node;
    }
    
    public TreeNode Find(TreeNode p, Stack st){
        
        while(!st.isEmpty()&&st.peek().val

对。。根本不用分类讨论

if (root == null || p == null) return null;
        TreeNode res=null;
        while(root!=null){
            if(root.val<=p.val)
                root=root.right;
            else{
                res=root;
                root=root.left;
            }
        }
        return res;

递归求

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

这个方法也可以求前面那个节点~改一下就好

public TreeNode predecessor(TreeNode root, TreeNode p) {
  if (root == null)
    return null;

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