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LeetCode OJ 99. Recover Binary Search Tree
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?
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解答:
这道题其实就是利用有错位时,中序遍历过程中会出现前后两个节点值的大小关系不符合BST性质的情况出现,这样只要把两个点都标记下来就好了……不过这里要注意1,2,3,4,5和1,3,2,4,5这样的错误只会在遍历中只会遇到一次错误情况,而1,2,3,4,5和3,2,1,4,5这样的错误在遍历中就会产生两次错误情况,所以在每一次遇到错误情况时就要标记两个点,如果有第二次遇到错误情况只要更新其中的一个标记就好了,当然要注意null节点……这个是用Java写的,用C写我不会啊TAT
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { TreeNode tmp = null, mis_node_1 = null, mis_node_2; void find_mistake(TreeNode root) { if(root==null) return; find_mistake(root.left); if(tmp != null&&root.val < tmp.val){ if(null == mis_node_1) mis_node_1 = tmp; mis_node_2 = root; } tmp = root; find_mistake(root.right); } public void recoverTree(TreeNode root) { find_mistake(root); if(null != mis_node_1){ int tmp_val = mis_node_1.val; mis_node_1.val = mis_node_2.val; mis_node_2.val = tmp_val; } } }
LeetCode OJ 99. Recover Binary Search Tree
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