《Cracking the Coding Interview》——第4章：树和图——题目5

2014-03-19 04:11

题目：设计算法检查一棵二叉树是否为二叉搜索树。

解法：既然是二叉搜索树，也就是说左子树所有节点都小于根，右子树所有节点都大于根。如果你真的全都检查的话，那就做了很多重复工作。只需要将左边最靠右，和右边最靠左的节点和根进行比较，然后依照这个规则递归求解即可。

代码：

// 4.5 Check if a binary tree is binary search tree.
#include <cstdio>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    
    TreeNode(int _val = 0): val(_val), left(nullptr), right(nullptr) {};
};

void constructBinaryTree(TreeNode *&root)
{
    int val;
    
    scanf("%d", &val);
    if (val <= 0) {
        root = nullptr;
    } else {
        root = new TreeNode(val);

        constructBinaryTree(root->left);
        constructBinaryTree(root->right);
    }
}

bool postorderTraversal(TreeNode *root, TreeNode *&left_most, TreeNode *&right_most)
{
    TreeNode *ll, *lr, *rl, *rr;
    bool res_left = true, res_right = true;
    
    if (root->left != nullptr) {
        if (!postorderTraversal(root->left, ll, lr)) {
            return false;
        }
        if (lr->val >= root->val) {
            // all left nodes must be smaller than the root.
            return false;
        }
    } else {
        ll = lr = root;
    }
    
    if (root->right != nullptr) {
        if (!postorderTraversal(root->right, rl, rr)) {
            return false;
        }
        if (rl->val <= root->val) {
            // all right nodes must be greater than the root.
            return false;
        }
    } else {
        rl = rr = root;
    }
    left_most = ll;
    right_most = rr;
    
    return true;
}

void clearBinaryTree(TreeNode *&root) {
    if (root == nullptr) {
        return;
    } else {
        clearBinaryTree(root->left);
        clearBinaryTree(root->right);
        delete root;
        root = nullptr;
    }
}

int main()
{
    TreeNode *root;
    TreeNode *left_most, *right_most;
    
    while (true) {
        constructBinaryTree(root);
        if (root == nullptr) {
            break;
        }
        
        left_most = right_most = nullptr;
        if (postorderTraversal(root, left_most, right_most)) {
            printf("Yes
");
        } else {
            printf("No
");
        }
        
        clearBinaryTree(root);
    }
    
    return 0;
}