LeetCode刷题笔录Recover Binary Search Tree
2014-08-20 00:08
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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?
一看到BST的问题,第一反应就应该是inorder traversal.利用O(n)的空间复杂度的话比较简单,用一个数组存储inorder traversal的所有结果,排个序,再赋值回去就好了。这个方法对于有任意N个元素位置不对的BST也可行。
这里只有两个node位置错了,因此可以只要找到这两个节点,用O(lgn)空间复杂度的递归就可以找到。维护三个指针,pre, first和second。pre用来存储inorder traversal的前一个节点,用以和当前节点进行比较来决定当前节点是否违反bst的性质。first和second存储两个放错了的node。
比如这个正确的BST:
1 2 3 4 5 6 7
交换一下1和4得到:
4 2 3 1 5 6 7
遍历时,pre=4,root=2时会发生第一次Inconsistency,这时让first=pre,即first=4.
第二次pre=3,root=1时会发生第二次inconsistency,这时让second=root,即second=1.
这样就找到了两个节点,交换他们的值就行了。
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?
一看到BST的问题,第一反应就应该是inorder traversal.利用O(n)的空间复杂度的话比较简单,用一个数组存储inorder traversal的所有结果,排个序,再赋值回去就好了。这个方法对于有任意N个元素位置不对的BST也可行。
这里只有两个node位置错了,因此可以只要找到这两个节点,用O(lgn)空间复杂度的递归就可以找到。维护三个指针,pre, first和second。pre用来存储inorder traversal的前一个节点,用以和当前节点进行比较来决定当前节点是否违反bst的性质。first和second存储两个放错了的node。
比如这个正确的BST:
1 2 3 4 5 6 7
交换一下1和4得到:
4 2 3 1 5 6 7
遍历时,pre=4,root=2时会发生第一次Inconsistency,这时让first=pre,即first=4.
第二次pre=3,root=1时会发生第二次inconsistency,这时让second=root,即second=1.
这样就找到了两个节点,交换他们的值就行了。
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { TreeNode pre; TreeNode first; TreeNode second; public void recoverTree(TreeNode root) { pre = null; first = null; second = null; inorder(root); int temp = first.val; first.val = second.val; second.val = temp; } public void inorder(TreeNode root){ if(root == null) return; inorder(root.left); if(pre == null){ pre = root; } else{ //a violation of BST if(pre.val > root.val){ if(first == null) first = pre; second = root; } pre = root; } inorder(root.right); } }如果要O(1)的空间复杂度的话,需要用到Threaded binary tree,可以不需要额外的stack来遍历二叉树。
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