balanced-binary-tree

gistfile1.txt

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

/*
solution: modified preorder (O(n))
매 순회마다 left, right 의 depth를 비교해서 >1 이면 return false 
more than 1 이라고 했으므로 인터뷰어와   
    1   와 같은 예제도 balanced 인지 사전 논의 필요
    / 
   2

1번에 accepted(tree, BST에 강한듯)
*/

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode *root) {
        if(root == NULL) return true;
        
        int leftDepth = findDepth(root->left, 1);
        int rightDepth = findDepth(root->right, 1);
        
        if(leftDepth == -1 || rightDepth == -1 || abs(leftDepth-rightDepth) > 1) {
            return false;
        }
        
        return true;
    }
    
    int findDepth(TreeNode* node, int depth) {
        if(node == NULL) {
            return depth;
        }
        
        int leftDepth = findDepth(node->left, depth+1);
        int rightDepth = findDepth(node->right, depth+1);
        
        if(leftDepth == -1 || rightDepth == -1 || abs(leftDepth-rightDepth) > 1) {
            return -1;
        }
        
        return max(leftDepth, rightDepth);
    }
};