【LeetCode】C# 98、Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node’s key.

The right subtree of a node contains only nodes with keys greater than the node’s key.

Both the left and right subtrees must also be binary search trees.

Example 1:

2

/ \

1 3

Binary tree [2,1,3], return true.

Example 2:

1

/ \

2 3

Binary tree [1,2,3], return false.

判断BST是否合法。

思路一：先按BST规则遍历root到List，然后判断是否合法。

这样比较繁琐，但是避免了一个问题。根节点和左右子节点比的话是没有问题的，但BST的定义是，何为二分查找树？1) 左子树的值都比根节点小；2) 右子树的值都比根节点大；3) 左右子树也必须满足上面两个条件。所以这是不行的。思路二就没考虑这个问题。

5

| \

4 10

/ \

3 11

/**
* Definition for a binary tree node.
* public class TreeNode {
*     public int val;
*     public TreeNode left;
*     public TreeNode right;
*     public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
List<int> list = new List<int>();
public bool IsValidBST(TreeNode root) {
if (root == null) return true;
if (root.left == null && root.right == null) return true;
inOrderTraversal(root);
for (int i = 1; i < list.Count; i++) {
if (list[i] <= list[i - 1]) return false;
}
return true;
}

public void inOrderTraversal(TreeNode root) {
if (root == null) return;
inOrderTraversal(root.left);
list.Add(root.val);
inOrderTraversal(root.right);
}
}

思路二：

if (root == null) return true;
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode pre = null;
while (root != null || stack.Count()!=0) {
while (root != null) {
stack.Push(root);
root = root.left;
}
root = stack.Pop();
if(pre != null && root.val <= pre.val) return false;
pre = root;
root = root.right;
}
return true;