20150624 lintcode 总结 Balanced Binary Tree

最新推荐文章于 2022-10-17 01:40:07 发布

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本文介绍了一种判断二叉树是否平衡的方法，并提供了一个具体的实现示例。平衡二叉树是指对于任意节点，其左右子树的高度差不超过1的二叉树。文中还给出了一些示例来帮助理解。

判断是否为balanced binary tree：

http://www.lintcode.com/en/problem/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.

Have you met this question in a real interview?

Yes

Example

Given binary tree A={3,9,20,#,#,15,7}, B={3,#,20,15,7}

A)  3            B)    3 
   / \                  \
  9  20                 20
    /  \                / \
   15   7              15  7

The binary tree A is a height-balanced binary tree, but B is not.

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    public boolean isBalanced(TreeNode root) {
        // write your code here
        if(root==null){
    		return true;
    	}
    	if(isBalanced(root.left) && isBalanced(root.right)){
    		int diff = Math.abs(depth(root.left)-depth(root.right));
    		if(diff<2){
    			return true;
    		}
    	}
    	return false;  		
    }
    
    private int depth(TreeNode root){
    	if(root==null){
    		return 0;
    	}
    	return 1+Math.max(depth(root.left), depth(root.right));   	
    }     

}