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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.
Inorder result is an ascending array for BST. Thus, this problem can be transferred to find successor for a node in BST.
successor = first element that is greater than target node.
Consider two situations:
1. Target node has right child
2. Target node doesn‘t have right child
Trace back, find first node whose left child is in the path from target node to root.
For this situation:
1. If node has parent pointer, we just use parent pointer.
2. If node doesn‘t have parent pointer, we use stack.
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public TreeNode inorderSuccessor(TreeNode root, TreeNode p) { 12 // Same question with finding successor in a BST 13 // 1. If p has right child, find minimum child in right subtree 14 // 2. If p doesn‘t have right child, find first ancestor whose left child is in the path from p to root. 15 TreeNode result = null; 16 if (p.right != null) { 17 result = p.right; 18 while (result.left != null) { 19 result = result.left; 20 } 21 } else { 22 Deque<TreeNode> stack = new LinkedList<TreeNode>(); 23 TreeNode current = root; 24 // push to stack 25 while (current != p && current != null) { 26 stack.push(current); 27 if (p.val < current.val) { 28 current = current.left; 29 } else { 30 current = current.right; 31 } 32 } 33 // pop stack 34 TreeNode prev = p; 35 while (!stack.isEmpty()) { 36 TreeNode cur = stack.pop(); 37 if (cur.left == prev) { 38 result = cur; 39 break; 40 } 41 prev = cur; 42 } 43 } 44 return result; 45 } 46 }
Solution 2 -- Recursive
Find Successor & Predecessor in BST
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原文地址：http://www.cnblogs.com/ireneyanglan/p/4946830.html
