Recover Binary Search Tree(回复二叉搜索树)
Recover Binary Search Tree(恢复二叉搜索树)
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?
题目原型:
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?
基本思路:
这个题的意思就是一个二叉查找树有两个元素被调换了顺序,此时需要修复成二叉查找树。基于此,由于构成二叉查找树的序列是二叉树的中序遍历序列,所以一旦我们根据中序遍历找到两个“变动”的点然后交换他们的值即可。
TreeNode pre = null;//始终指向中序遍历的上一个节点 TreeNode n1 = null;//记录待交换节点 TreeNode n2 = null;//记录带交换节点 public TreeNode recoverTree(TreeNode root) { findTwoNode(root); if(n1!=null&&n2!=null) { int tmp = n1.val; n1.val = n2.val; n2.val = tmp; } return root; } public void findTwoNode(TreeNode root) { if(root==null) return; findTwoNode(root.left); if(pre!=null&&pre.val>root.val) { if(n1==null) n1 = pre; n2 = root; } pre = root; findTwoNode(root.right); }