36.平衡二叉树(273)

• 题目描述：输入一棵二叉树，判断该二叉树是否是平衡二叉树。

  若一颗二叉树中每一个结点的左、右子树的高度最多相差1，则称此二叉树为平衡二叉树。

• 思路：

• 代码：

  package _36.平衡二叉树;
  
  /**
   * 题目描述：输入一棵二叉树，判断该二叉树是否是平衡二叉树。
   * 
   * @author Administrator
   *
   */
  public class BalancedBTree {
  	public boolean IsBalanced_Solution(TreeNode root) {
  		if(root == null) return true;
  		//左子树的高度
  		int left  = depth(root.left);
  		//右子树的高度
  		int right = depth(root.right);
  		//左右子树的高度差
  		int differ = left - right;
  		if(differ > 1 || differ < -1){
  			return false;
  		}
  		return IsBalanced_Solution(root.left) && IsBalanced_Solution(root.right); 
  	}
  	/**
  	 * 求树的高度
  	 * @param root
  	 * @return
  	 */
  	public int depth(TreeNode root){
  		if(root == null) return 0;
  		int left,right;
  		left  = depth(root.left) + 1;
  		right = depth(root.right) + 1;
  		return left > right ? left:right;
  	}
  	
  	
  	public static void main(String[] args) {
  		/**
  		 *          5
  			      /   \
  			     3     7
  			    / \   
  			   2   4 
  			  / \
  			 6   8
  		 */   
  		TreeNode root = new TreeNode(5);
          root.left = new TreeNode(3);
          root.left.left = new TreeNode(2);
          root.left.right = new TreeNode(4);
          root.right = new TreeNode(7);
          root.left.left.left = new TreeNode(6);
          root.left.left.right = new TreeNode(8);
          
          boolean isBalanced_Solution = new BalancedBTree().IsBalanced_Solution(root);
          System.out.println(isBalanced_Solution);
  	}
  }
  
  class TreeNode {
  	int val = 0;
  	TreeNode left = null;
  	TreeNode right = null;
  
  	public TreeNode(int val) {
  		this.val = val;
  	}
  }