【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
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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
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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;