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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?
ref
http://www.cnblogs.com/springfor/p/3891390.html
1 / 5 3 / 4 2
just like in the ref" 所以在中序便利时，遇见的第一个顺序为抵减的两个node，大的那个肯定就是要被recovery的其中之一，要记录。"
in order search 的缘故是因为这样可以保证左面能先被traverse到，就符合了找到第一个swap点的目的
public class Solution { TreeNode first, second, pre; public void recoverTree(TreeNode root) { if(root==null) return; inorder(root); if(first!=null && second!=null){ int tmp=first.val; first.val = second.val; second.val = tmp; } return; } private void inorder(TreeNode node){ if(node==null) return; inorder(node.left); if(pre==null){ pre=node; }else{ if(pre.val>node.val){ if(first==null){ first=pre; } second = node; //keep moving on to find the second on the right side } pre = node; } inorder(node.right); } }
天题系列： Recover Binary Search Tree
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原文地址：http://www.cnblogs.com/jiajiaxingxing/p/4565038.html
