110.Balanced Binary Tree¶

110.Balanced Binary Tree¶

Tags: Tree Depth-first Search Easy

Links: https://leetcode.com/problems/balanced-binary-tree/

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

a binary tree in which the left and right subtrees of every node differ in height by no more than 1.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

Return true. Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

Return false.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include <cmath>
class Solution {
public:
    bool isBalanced(TreeNode* root) {
        if (!root) return true;

        if (isBalanced(root -> left) && isBalanced(root -> right)) {
            int leftHeight = height(root -> left);
            int rightHeight = height(root -> right);
            if (abs(leftHeight - rightHeight) > 1) return false;
            else return true;
        }

        return false;
    }

    int height(TreeNode *root) {
        if (!root) return 0;
        if (!root -> left && !root -> right) return 1;
        return max(height(root -> left), height(root -> right)) + 1;
    }
};

非常容易写出来的递归写法。但是还可以进行优化，因为我们每次计算高度都要遍历此节点以下的节点，实际上如果此节点的子树不是平衡的，那么无须再检查了。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include <cmath>
class Solution {
public:
    bool isBalanced(TreeNode* root) {        
        if (checkDepth(root) == -1) return false;

        return true;
    }

    int checkDepth(TreeNode *root) {
        if (!root) return 0;
        int leftHeight = checkDepth(root -> left);
        if (leftHeight == -1) return -1;
        int rightHeight = checkDepth(root -> right);
        if (rightHeight == -1) return -1;

        if (abs(leftHeight - rightHeight) > 1) return -1;
        return max(leftHeight, rightHeight) + 1;
    }
};

如果我们发现子树不平衡，则不计算具体的深度，而是直接返回-1。那么优化后的方法为：对于每一个节点，我们通过checkDepth函数递归获得左右子树的深度，如果子树是平衡的，则返回真实的深度，若不平衡，直接返回-1，此方法时间复杂度O(N)，空间复杂度O(H)