[LeetCode] 110. Balanced Binary Tree 平衡二叉树

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.

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.

给定一个二叉树,判断是否高度平衡。高度平衡二叉树的定义:二叉树的任意节点的两个子树的深度差不超过1。

解法:根据定义,只需要判定一颗二叉树的左右子树高度的高度差是否小于等于1。递归处理每一颗二叉树左右子树的高度,并进行判断再回溯。

Java:

class Solution {
    int abs(int x) {
        return x > 0 ? x : -x;
    }
    
    int check(TreeNode* root) {
       if (!root) return NULL;
       
       int lch = check(root -> left);
       int rch = check(root -> right);
       // 检查子树是否存在不平衡
       if (lch == -1 || rch == -1 || abs(lch - rch) > 1) return -1;
       
       // 返回当前子树高度
       return (lch > rch ? lch : rch) + 1;
    }
public:
    bool isBalanced(TreeNode* root) {
        return check(root) != -1;
    }
};  

Java:  without ResultType

public class Solution {
    public boolean isBalanced(TreeNode root) {
        return maxDepth(root) != -1;
    }

    private int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }

        int left = maxDepth(root.left);
        int right = maxDepth(root.right);
        if (left == -1 || right == -1 || Math.abs(left-right) > 1) {
            return -1;
        }
        return Math.max(left, right) + 1;
    }
} 

Python:

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    # @param root, a tree node
    # @return a boolean
    def isBalanced(self, root):
        return (self.getHeight(root) >= 0)
    
    def getHeight(self, root):
        if root is None:
            return 0
        left_height, right_height = self.getHeight(root.left), self.getHeight(root.right)
        if left_height < 0 or right_height < 0 or abs(left_height - right_height) > 1:
            return -1
        return max(left_height, right_height) + 1 

Python:

class Solution:
    """
    @param root: The root of binary tree.
    @return: True if this Binary tree is Balanced, or false.
    """
    def isBalanced(self, root):
        balanced, _ = self.validate(root)
        return balanced
        
    def validate(self, root):
        if root is None:
            return True, 0
            
        balanced, leftHeight = self.validate(root.left)
        if not balanced:
            return False, 0
        balanced, rightHeight = self.validate(root.right)
        if not balanced:
            return False, 0
            
        return abs(leftHeight - rightHeight) <= 1, max(leftHeight, rightHeight) + 1  

C++:

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));
    }
};

C++:

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);
    }
};

C++:

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    int depth(TreeNode *root) {
        if (root == NULL) {
            return 0;
        }
        int left = depth(root->left);
        int right = depth(root->right);
        if (left == -1 || right == -1 || abs(left - right) > 1) {
            return -1;
        }
        return max(left, right) + 1;
    }

    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    bool isBalanced(TreeNode *root) {
        return depth(root) != -1;
    }
};

  

   

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posted @ 2018-03-22 09:06  轻风舞动  阅读(582)  评论(0编辑  收藏  举报