lintcode balanced-binary-tree 平衡二叉树

问题描述

平衡二叉树

笔记

这个题是没想出来，最后参考了九章算法的解法，是最高效的，把depth函数做一点修改，如果不平衡的情况下返回-1。（代码1）

代码2是剑指offer的重复遍历子节点的程序。理解简单，效率最低。

代码3是代码2的改进版，不需要重复遍历节点。“一边遍历，一边记录深度”。

代码1

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {public:    /**     * @param root: The root of binary tree.     * @return: True if this Binary tree is Balanced, or false.     */    bool isBalanced(TreeNode *root) {        // write your code here        return depth(root) != -1;    }    int depth(TreeNode *root)    {        if (root == NULL)            return 0;        int left = depth(root->left);        int right = depth(root->right);        if (left == -1 || right == -1 || abs(left-right) > 1)            return -1;        return max(left, right) + 1;    }};

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {public:    /**     * @param root: The root of binary tree.     * @return: True if this Binary tree is Balanced, or false.     */    int depth(TreeNode *root)    {        if (root == NULL)            return 0;        return max(depth(root->left), depth(root->right)) + 1;    }    bool isBalanced(TreeNode *root) {        // write your code here        if (root == NULL)            return true;        if (!isBalanced(root->left))            return false;        if (!isBalanced(root->right))            return false;        int left = depth(root->left);        int right = depth(root->right);        if (abs(left-right) > 1)            return false;        return true;    }};

代码3

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {public:    /**     * @param root: The root of binary tree.     * @return: True if this Binary tree is Balanced, or false.     */    bool balance(TreeNode *root, int &depth)    {        if (root == NULL)        {            depth = 0;            return true;        }        int left, right;        if (balance(root->left, left) && balance(root->right, right))        {            if (abs(left - right) <= 1)            {                depth = max(left, right) + 1;                return true;            }        }        return false;    }    bool isBalanced(TreeNode *root) {        // write your code here        int depth = 0;        return balance(root, depth);    }};