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);
    }
};