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  • LeetCode | Balanced Binary Tree

    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.

    OJ's Binary Tree Serialization:    //LeetCode对二叉树的表示方法

    The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

    Here's an example:

       1
      / 
     2   3
        /
       4
        
         5
    

    The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        
        public int maxDepth(TreeNode root){        //此方法为10.26日的题,递归求树的最大深度
               if(root==null) return 0;
               int leftDepth = maxDepth(root.left);
               int rightDepth = maxDepth(root.right);
               int depth =  leftDepth>=rightDepth?leftDepth:rightDepth ; 
               return depth+1;
           }
           
        public boolean isBalanced(TreeNode root) {
            if (root==null) return true;               //注意,空树也认为是平衡树的情况
    
           int leftHeight = maxDepth(root.left);       //求左右子树的高度
           int rightHeight = maxDepth(root.right);
          
           if( Math.abs(leftHeight-rightHeight)>1 ) return false;  //左右子树的高度相差大于1,一定不是平衡树
           return isBalanced(root.left)&&isBalanced(root.right);   //注意!!!此处不能直接返回FALSE,要继续向下判断,
        }                                                          //直至确认每个节点都满足平衡的条件才返回true
        
    }
    //左右子树的高度差不大于一,并不代表一定是平衡树,平衡树要求每个节点的左右子节点高度差都不大于1,不只是root节点
    //例如:树{1 2 2 3 # # 3 4 # # 4},其根的左右子树高度一样,但其不是平衡树(#代表此节点为null)



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  • 原文地址:https://www.cnblogs.com/dosmile/p/6444473.html
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