判断是否为二叉搜索树

最新推荐文章于 2020-07-06 16:01:08 发布

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本文介绍了一种判断二叉树是否为有效二叉搜索树的方法。通过中序遍历确保节点值递增，或者使用递归方式，为每个节点设定合理的值范围，从而判断整个二叉树是否符合二叉搜索树的定义。

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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:

Input:
    2
   / \
  1   3
Output: true

Example 2:

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

/**

* 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;
}
if(root->left==NULL&&root->right==NULL){
return true;
}

TreeNode *cur=root,*pre=NULL;

stack<TreeNode *> st; // 中序遍历
while(!st.empty()||cur){

while(cur){
st.push(cur);
cur=cur->left;
}

if(!st.empty()){
cur=st.top();
st.pop();

// ivec.push_back(cur->val)
if(pre==NULL){
pre=cur;
}else{
if(cur->val<=pre->val){
return false;
}else{
pre=cur;
}
}

cur=cur->right;
}


}
return true;


// return helper1(root,INT_MIN+1,INT_MAX-1);

}


bool helper1(TreeNode *root,int low,int upper){
if(root==NULL){
return true;
}
if(root->val<=low||root->val>=upper){// 二查搜索数不能有等于
return false;
}

return helper(root->left,low,root->val)&&helper(root->right,root->val,upper);

}

// 目标找出一个对应的 low 和 upper 与 root->val进行比较
bool helper(TreeNode *root,int low,int upper){
if(root==NULL){
return true;
}
if(root->val>=upper||root->val<=low){
return false;
}
// 往左走最大值就会变化往右走最小值就会变化
return helper(root->left,low,root->val)&&helper(root->right,root->val,upper);
}
};