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

Input: [2,1,3]
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

A:

-------利用传入的参数，每次返回其子树的范围--------再对比-------------

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        int minV =0, maxV = 0;
        if(!root)
            return true;
        return helper(root, minV, maxV);
    }
private:
    // return whether it is the BST, setting maxLeft, and minRight of the tree rooted at root
    bool helper(TreeNode* root, int &minVal, int &maxVal)
    {// root is not NULL        
        minVal = root->val, maxVal = root->val;
        if(root->left)
        {
            int a,b;
            if( ! helper(root->left, a, b))
                return false;
            
            if( !( root->val > b) )
                return false;
            
            minVal = a;
        }
        if(root->right)
        {
            int c, d;
            if(!helper(root->right, c, d))
                return false;
            
            if( ! (root->val < c ) )
                return false;
            maxVal = d;
        }
        return true;
    }
};

------------------------不同于每次返回其子树的范围， 我们给出其范围。----但是这样， 如果val 是double就会有问题-------------------

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    bool isValidBST(TreeNode* root) {        
        return helper(root, LONG_MIN, LONG_MAX);
    }
private:
    bool helper(TreeNode* root, long low, long high)
    {
        return !root || \
            (root->val > low && \
             root->val < high && \
             helper(root->left,  low, root->val) && \
             helper(root->right, root->val, high));
    }
};

-----------------------------把树弄成一个list, 然后对比。---------------------

public class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root == null){
            return true;
        }
        // in-order traversal, to compare
        Stack<Integer> stack = new Stack<Integer>();
        inOrderTraversal(root,stack);
        //stack is not empty, since root is not null
        int lastPop = stack.pop();
        while(!stack.isEmpty()){
            int curPop = stack.pop();
            if(curPop >= lastPop){
                return false;
            }
            lastPop = curPop;
        }
        return true;
    }
    private void inOrderTraversal(TreeNode root, Stack<Integer> stack){
        if(root == null){
            return;
        }
        inOrderTraversal(root.left,stack);
        stack.push(root.val);
        inOrderTraversal(root.right,stack);
    }
}

Mistakes:
1:刚开始，没注意，是strictly less than,  (more than)
        哎，第二遍，又是犯了这个错误。
2:  这道题目的关键点就是  root.val可能是极值。因此要跟Long 的值对比。