98. Validate Binary Search Tree (Medium)

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

A valid BST is defined as follows:

• The left  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:

Input: root = [2,1,3]
Output: true

Example 2:

Input: root = [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.

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -231 <= Node.val <= 231 - 1

A:

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

------------------------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() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) { return helper(LONG_MIN, root, LONG_MAX); }
private:
    bool helper(long lower, TreeNode* root, long upper) {
        if (!root)
            return true;
        long val = root->val;
        return val > lower && val < upper && helper(lower, root->left, val) &&
               helper(val, root->right, upper);
    }
};

-----------------------------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。