110. Balanced Binary Tree

最新推荐文章于 2022-07-18 20:53:17 发布

转载最新推荐文章于 2022-07-18 20:53:17 发布 · 54 阅读

本文介绍了一种用于判断二叉树是否为高度平衡的方法。通过递归算法获取每个节点的高度，并确保任一节点的左右子树高度差不超过1。提供了两种实现方式，一种直接比较左右子树高度，另一种采用辅助函数进行深度计算。

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110. Balanced Binary Tree

题目

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 every node never differ by more than 1.

解析

class Solution_110 {
public:
    int getHeight(TreeNode* root)
    {
         if (!root)
         {
             return 0;
         }

         int left = getHeight(root->left);
         int right = getHeight(root->right);

         return (left > right) ? (left + 1):(right + 1);
    }

    bool isBalanced(TreeNode *root) {
        if (!root)
        {
            return true;
        }
        // 不应该用<=1,return true,这样就提前返回了 
        //if (abs(getHeight(root->left) - getHeight(root->right)) > 1) {
        //  return false;
        //}
        //递归：自顶向下
        return  abs(getHeight(root->left) - getHeight(root->right) <= 1) && isBalanced(root->left) && isBalanced(root->right);
    }

    // method2
    bool isBalanced1(TreeNode *root) {
        int depth = 0;
        return isBalanced_helper(root, depth);
    }
    bool isBalanced_helper(TreeNode *root, int &depth) {
        if (root == NULL){
            depth = 0;
            return true;
        }
        int left, right;
        if (isBalanced_helper(root->left, left) && isBalanced_helper(root->right, right)){
            if (abs(left - right) <= 1){
                depth = 1 + max(left, right);
                return true;
            }
        }
        return false;
    }
};