Return the root node of a binary search tree that matches the given preorder
traversal.
(Recall that a binary search tree is a binary tree where for every node, any descendant of node.left
has a value <
node.val
, and any descendant of node.right
has a value >
node.val
. Also recall that a preorder traversal displays the value of the node
first, then traverses node.left
, then traverses node.right
.)
Example 1:
Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]
Note:
1 <= preorder.length <= 100
- The values of
preorder
are distinct.
题目理解:
根据给定的先序遍历(先访问根节点,后访问左右子节点)结果,重构二叉搜索树(左子节点的值小于根节点,右子节点的值大于根节点)
解题思路:
递归建立二叉搜索树。
由于是先序遍历,遍历结果的第一个值一定是根节点的值,然后是左子树遍历结果,最后是右子树遍历结果,左子树遍历序列中的每一个值一定都小于根节点的值,根据这个性质,可以找出左子树遍历结果和右子树遍历结果,然后递归创建左子树和右子树。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
int[] preorder;
public TreeNode build(int start, int end){
if(start > end)
return null;
TreeNode root = new TreeNode(preorder[start]);
int pos = start + 1;
while(pos <= end && preorder[start] > preorder[pos])
pos++;
root.left = build(start + 1, pos - 1);
root.right = build(pos, end);
return root;
}
public TreeNode bstFromPreorder(int[] preorder) {
this.preorder = preorder;
return build(0, preorder.length - 1);
}
}