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Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
1 /** 2 * Definition for binary tree 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 12 13 TreeNode n1 = null; 14 TreeNode n2 = null; 15 TreeNode pre = null; 16 17 public void recoverTree(TreeNode root) { 18 findNodes(root); 19 if (n1 != null && n2 != null) { 20 int temp = n1.val; 21 n1.val = n2.val; 22 n2.val = temp; 23 } 24 } 25 26 void findNodes(TreeNode root) { 27 if (root == null) { 28 return; 29 } 30 findNodes(root.left); 31 if (pre != null && pre.val > root.val) { 32 if (n1 == null) { 33 n1 = pre; 34 } 35 n2 = root; 36 } 37 pre = root; 38 findNodes(root.right); 39 } 40 41 42 }
LeetCode Recover Binary Search Tree
标签:des style blog http ar io color sp for
原文地址:http://www.cnblogs.com/birdhack/p/4172965.html