Assume a BST is defined as follows:
class Solution {
public:
bool isValidBST(TreeNode *root) {
return check(root, INT_MIN, INT_MAX);
}
private:
bool check(TreeNode *root, int left, int right){
if(root == NULL)
return true;
return (root->val > left) && (root->val < right)
&& check(root->left, left, root->val) &&check(root->right, root->val, right);
}//这里的左儿子的左界用上面传下来的,右界用节点值,右儿子镜面对称
};/**
* Definition for binary tree
* 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;
bool bleft = true, bright = true;
if (root->left != NULL){
if (root->val > root->left->val){ //注意这里检验以及递归是不能保证根节点大于左边所有的,矛盾在于,左儿子的右儿子,以及右儿子的左儿子;
bleft = isValidBST(root->left);
}
else{
return false;
}
}
if (root->right != NULL){
if (root->val < root->right->val){
bright = isValidBST(root->right);
}
else{
return false;
}
}
return bleft && bright;
}
};LeetCode :: Validate Binary Search Tree[详细分析],布布扣,bubuko.com
LeetCode :: Validate Binary Search Tree[详细分析]
原文地址:http://blog.csdn.net/u013195320/article/details/37672351