98. Validate Binary Search Tree (Tree; DFS)

 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.

思路：由于Binary Tree的中序遍历结果是正序，所以可以检查中序遍历的结果是否递增

/**
 * 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) {
        if(root==NULL) return true;
        TreeNode* pre = NULL; //we don't need to save all nodes, only a previous node is enough to know whether it's an increase sequence
        return inOrderTraverse(root,pre);
    }
    
    bool inOrderTraverse(TreeNode* root, TreeNode* &pre){ //important to use &, otherwise new object will use a new address and the result won't bring back to caller
        //visit left child
        if(root->left) 
            if(!inOrderTraverse(root->left,pre))
                return false;
        
        //visit root
        if(pre==NULL) pre = new TreeNode(root->val);
        else if(root->val > pre->val) pre->val = root->val;
        else return false;
        
        //visit right child
        if(root->right) return inOrderTraverse(root->right, pre);
        else return true;
    }
};