[LeetCode 题解]: Validate Binary Search Tree
2014-07-31 13:37
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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.
confused what
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
The above binary tree is serialized as
题意:给定一颗二叉树,判断该树是否是二叉排序树。
二叉排序树的规定:
(1)如果存在左子树,则左子树的所有节点的值都要小于根节点的值。
(2)如果存在右子树,则右子树的所有节点的值都要大于根节点的值。
(3)如果一棵以root为根节点的树是二叉排序树,那么其左子树和右子树也必须是二叉排序树。
先给出我想到的解:
在之前有写到过用队列创建一颗二叉排序树,其过程类似于二叉树的中序遍历。
因此,可以反过来看出二叉排序树的中序遍历的结果组成的序列应该是有序的。
根据这种想法,可以得到如下代码:
转载请注明出处: http://www.cnblogs.com/double-win/ 谢谢!
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.
confused what
"{1,#,2,3}"means?
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1 / \ 2 3 / 4 \ 5
The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}".
题意:给定一颗二叉树,判断该树是否是二叉排序树。
二叉排序树的规定:
(1)如果存在左子树,则左子树的所有节点的值都要小于根节点的值。
(2)如果存在右子树,则右子树的所有节点的值都要大于根节点的值。
(3)如果一棵以root为根节点的树是二叉排序树,那么其左子树和右子树也必须是二叉排序树。
先给出我想到的解:
在之前有写到过用队列创建一颗二叉排序树,其过程类似于二叉树的中序遍历。
因此,可以反过来看出二叉排序树的中序遍历的结果组成的序列应该是有序的。
根据这种想法,可以得到如下代码:
class Solution { public: bool isValidBST(TreeNode *root) { if(!root ||(!root->left && !root->right)) return true; vector<int> vi; vi.clear(); // 用非递归的方式对树进行中序遍历,将结果存放到vi数组中 stack<TreeNode* > s; TreeNode *tmp=root; while(!s.empty() || tmp){ if(tmp){ s.push(tmp); tmp = tmp->left; }else{ tmp = s.top(); s.pop(); vi.push_back(tmp->val); tmp = tmp->right; } } //对中序遍历的结果进行判断,注意不能有重复的数字 for(int i=1;i<vi.size();i++){ if(vi[i]<=vi[i-1]) return false; } return true; } };
转载请注明出处: http://www.cnblogs.com/double-win/ 谢谢!
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