897. 递增顺序查找树

港控/mmm° 2023-10-05 21:09 34阅读 0赞

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。

**示例 ：**

输入：[5,3,6,2,4,null,8,1,null,null,null,7,9]
    
           5
          / \
        3    6
       / \    \
      2   4    8
     /        / \ 
    1        7   9
    
    输出：[1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
    
     1
      \
       2
        \
         3
          \
           4
            \
             5
              \
               6
                \
                 7
                  \
                   8
                    \
                     9

**提示：**

1.  给定树中的结点数介于 `1` 和 `100` 之间。
2.  每个结点都有一个从 `0` 到 `1000` 范围内的唯一整数值。

/**
     * 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) {
    		Stack<TreeNode> stack = new Stack<>();
    		ArrayList<Integer> al = new ArrayList<>();
    		TreeNode curr = root;
    		while (!stack.isEmpty() || curr != null) {
    			while (curr != null) {
    				stack.push(curr);
    				curr = curr.left;
    			}
    			TreeNode node = stack.pop();
    			al.add(node.val);
    			curr = node.right;
    		}
    		if (al != null) {
    			TreeNode node = new TreeNode(al.get(0));
    			TreeNode head = new TreeNode(0, null, node);
    			for (int i = 1; i < al.size(); i++) {
    				node.right = new TreeNode(al.get(i));
    				node = node.right;
    			}
    			return head.right;
    		}
    		return null;
    
    	}
    }