算法第八周解题报告

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98. Validate Binary Search Tree

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.

解题思路：本人的思路是通过后序遍历，对于每个非叶子结点:

1.判断他们的左右子树是否是搜索二叉树

2.取回左右子树的最大最小值，判断以这个节点为根的树是否是搜索二叉树。

算法代码如下:

/** * 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 max, min;        return travel(root, min, max);    }    bool travel(TreeNode* root, int & min, int & max){        if (root == NULL) {            max = INT_MIN;            min = INT_MAX;            return true;        }        int leftMin, leftMax, rMin, rMax;        bool isBSTree = travel(root -> left, leftMin, leftMax);        if(isBSTree == false) return false;        isBSTree = travel(root -> right, rMin, rMax);        if(isBSTree == false)return false;        if(root -> left != NULL && root -> val <= leftMax) return false;        if(root -> right != NULL && root -> val >= rMin) return false;        max = leftMax > rMax ? leftMax : rMax;        max = root -> val > max ? root -> val : max;        min = leftMin < rMin ? leftMin : rMin;        min = root -> val < min ? root -> val : min;        return true;    }};

结果如下:

0 0