[LeetCode] 897. Increasing Order Search Tree

Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.

Example 1:

Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

Example 2:

Input: root = [5,1,7]
Output: [1,null,5,null,7]

Constraints:
The number of nodes in the given tree will be in the range [1, 100].
0 <= Node.val <= 1000

递增顺序搜索树。

给你一个树，请你 按中序遍历 重新排列树，使树中最左边的结点现在是树的根，并且每个结点没有左子结点，只有一个右子结点。

思路

这道题跟之前的114题很像，也是属于需要把树做扁平化处理的题目。既然题目都说了用中序遍历，我这里给出两种做法，迭代和递归。时间空间复杂度均是 O(n)。看代码应该能明白思路，只是需要注意当处理节点的时候，需要找到新的 head 节点，以及把处理过的每个节点的左指针设置成 null，新的树里面每个节点都只有右指针。

复杂度

时间O(n)
空间O(n)

代码

Java迭代实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode increasingBST(TreeNode root) {
        // corner case
        if (root == null) {
            return null;
        }

        // normal case
        TreeNode dummy = new TreeNode(0); // 虚拟头节点
        TreeNode pre = dummy;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while (!stack.isEmpty() || cur != null) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            cur.left = null;
            pre.right = cur;
            pre = cur;
            cur = cur.right;
        }
        return dummy.right;
    }
}

Java递归实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
	TreeNode pre = null;
	TreeNode head = null;

	public TreeNode increasingBST(TreeNode root) {
		if (root == null) {
			return null;
		}
		increasingBST(root.left);
		if (head == null) {
			head = root;
		}
		if (pre != null) {
			root.left = null;
			pre.right = root;
		}
		pre = root;
		increasingBST(root.right);
		return head;
	}
}

相关题目

114. Flatten Binary Tree to Linked List
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