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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.
解法一：俺自个儿的方法，居然跑了16ms。。。妈蛋！
public class Solution { public TreeNode inorderSuccessor(TreeNode root, TreeNode p) { ArrayList<TreeNode> res = inOrderTrav(root); for(int i=0;i<res.size()-1;i++) { if(p==res.get(i)) return res.get(i+1); } return null; } public ArrayList<TreeNode> inOrderTrav(TreeNode root) { LinkedList<TreeNode> stack = new LinkedList<TreeNode>(); ArrayList<TreeNode> res = new ArrayList<TreeNode>(); if(root==null) return res; while(root!=null||!stack.isEmpty()) { if(root!=null) { stack.push(root); root = root.left; } else { root = stack.pop(); res.add(root); root = root.right; } } return res; } }
解法二：
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) { TreeNode succ = null; while (root != null) { if (p.val < root.val) { succ = root; root = root.left; } else root = root.right; } return succ; }
reference：https://leetcode.com/discuss/61105/java-python-solution-o-h-time-and-o-1-space-iterative
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原文地址：http://www.cnblogs.com/hygeia/p/5104279.html
