[LeetCode] Balanced Binary Tree 平衡二叉樹

 

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 depth of the two subtrees of everynode never differ by more than 1.

Example 1:

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

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

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

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

Return false.

 

求二叉樹是否平衡，根據題目中的定義，高度平衡二叉樹是每一個結點的兩個子樹的深度差不能超過1，那么我們肯定需要一個求各個點深度的函數，然后對每個節點的兩個子樹來比較深度差，時間復雜度為O(NlgN)，代碼如下：

 

解法一：

class Solution {
public:
    bool isBalanced(TreeNode *root) {
        if (!root) return true;
        if (abs(getDepth(root->left) - getDepth(root->right)) > 1) return false;
        return isBalanced(root->left) && isBalanced(root->right);    
    }
    int getDepth(TreeNode *root) {
        if (!root) return 0;
        return 1 + max(getDepth(root->left), getDepth(root->right));
    }
};

 

上面那個方法正確但不是很高效，因為每一個點都會被上面的點計算深度時訪問一次，我們可以進行優化。方法是如果我們發現子樹不平衡，則不計算具體的深度，而是直接返回-1。那么優化后的方法為：對於每一個節點，我們通過checkDepth方法遞歸獲得左右子樹的深度，如果子樹是平衡的，則返回真實的深度，若不平衡，直接返回-1，此方法時間復雜度O(N)，空間復雜度O(H)，參見代碼如下：

 

解法二：

class Solution {
public:    
    bool isBalanced(TreeNode *root) {
        if (checkDepth(root) == -1) return false;
        else return true;
    }
    int checkDepth(TreeNode *root) {
        if (!root) return 0;
        int left = checkDepth(root->left);
        if (left == -1) return -1;
        int right = checkDepth(root->right);
        if (right == -1) return -1;
        int diff = abs(left - right);
        if (diff > 1) return -1;
        else return 1 + max(left, right);
    }
};

 

類似題目：

Maximum Depth of Binary Tree

 

參考資料：

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

https://leetcode.com/problems/balanced-binary-tree/discuss/35691/The-bottom-up-O(N)-solution-would-be-better

https://leetcode.com/problems/balanced-binary-tree/discuss/35686/Java-solution-based-on-height-check-left-and-right-node-in-every-recursion-to-avoid-further-useless-search

 

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