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 Input: [2,1,3] Output: true
Example 2:
5 / \ 1 4 / \ 3 6 Input: [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.
A:
-------利用传入的参数,每次返回其子树的范围--------再对比-------------
/** * 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) { int minV =0, maxV = 0; if(!root) return true; return helper(root, minV, maxV); } private: // return whether it is the BST, setting maxLeft, and minRight of the tree rooted at root bool helper(TreeNode* root, int &minVal, int &maxVal) {// root is not NULL minVal = root->val, maxVal = root->val; if(root->left) { int a,b; if( ! helper(root->left, a, b)) return false; if( !( root->val > b) ) return false; minVal = a; } if(root->right) { int c, d; if(!helper(root->right, c, d)) return false; if( ! (root->val < c ) ) return false; maxVal = d; } return true; } };
------------------------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(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isValidBST(TreeNode* root) { return helper(root, LONG_MIN, LONG_MAX); } private: bool helper(TreeNode* root, long low, long high) { return !root || \ (root->val > low && \ root->val < high && \ helper(root->left, low, root->val) && \ helper(root->right, root->val, high)); } };
-----------------------------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。
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