Problem
Given a root of Binary Search Tree with unique value for each node. Remove the node with given value. If there is no such a node with given value in the binary search tree, do nothing. You should keep the tree still a binary search tree after removal.
ExampleGiven binary search tree:
5 / 3 6 / 2 4
Remove 3, you can either return:
5 / 2 6 4
or
5 / 4 6 / 2Note
以下是rebuild的示意:先让r到达r的最左端,然后将l接在r的左子树,这样就把所有比root.left大的结点都集合在root.right了。将root.right接在root.left的右子树,然后返回root.left即可。
(1)
root / left right / / x l r x
(2)
left right / / x r x / l
(3)
left / x right / r x / lSolution
public class Solution { public TreeNode removeNode(TreeNode root, int value) { TreeNode dummy = new TreeNode (-1), pre = dummy, cur = root; dummy.right = root; while (cur != null) { if (cur.val == value) { if(pre.left == cur) { pre.left = makenew(cur); } else pre.right = makenew(cur); break; } else if (cur.val < value) { pre = cur; cur = cur.right; } else { pre = cur; cur = cur.left; } } return dummy.right; } private TreeNode makenew(TreeNode node) { if (node.left == null || node.right == null) { return node.left == null ? node.right : node.left; } TreeNode left = node.left.right; TreeNode right = node.right.left; while (right.left != null) { right = right.left; } right.left = left; node.left.right = node.right; return node.left; } }
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