110. Balanced Binary Tree (easy)

Q:

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]:

    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.

A:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode* root) {
        int l=0;
        return helper(root, l);
    }
private:
    bool helper(TreeNode* root, int &depth)
    {        
        if(!root)
        {
            depth = 0;
            return true;
        }
        int l=0, r = 0;
        bool isLeft = helper(root->left, l);
        bool isRight = helper(root->right, r);
        depth = 1+ max(l, r);
        return isLeft && isRight && abs(l-r)<=1;
    }
};

Mistakes:


Learned:

---------------------Java,  因为不能引用传值，  ---------如果不平衡，就返回负 的树的高度

public class BalancedBinaryTree {
    public boolean isBalanced(TreeNode root) {
        return getHeight(root) >= 0 ;
    }

    private int getHeight(TreeNode root) {
        // get the weight, but if tree is unbalanced, return -1
        if (root == null)
            return 0;
        int left = getHeight(root.left);
        int right = getHeight(root.right);
        if (Math.abs(left - right) > 1 || left < 0 || right < 0)
            return -1;
        return 1 + Math.max(left, right);
    }
}