【复】判断树的平衡，

/**

 * Definition for binary tree

 * public class TreeNode {

 *     int val;

 *     TreeNode left;

 *     TreeNode right;

 *     TreeNode(int x) { val = x; }

 * }

 */

public class Solution {

    public boolean isBalanced(TreeNode root) {

        if(root==null) return true;

        int d1=depth(root.right);

        int d2=depth(root.left);

        if(Math.abs(d1-d2)>1) return false;

        else  return isBalanced(root.left)&&isBalanced(root.right);

         

        

        

        

    }

    

    public int depth(TreeNode node)

    {

        if(node==null) return 0;

        int d1=depth(node.right);

        int d2=depth(node.left);

        

        int max=d1>d2?d1:d2;

        return  max+1;

        

    }

    

}

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2.求高度的时候在不平衡的时候返回-1就ok了，不用求高度，但是效率好像没啥变化

/**

 * Definition for binary tree

 * public class TreeNode {

 *     int val;

 *     TreeNode left;

 *     TreeNode right;

 *     TreeNode(int x) { val = x; }

 * }

 */

public class Solution {

    private boolean ans=true;

    public boolean isBalanced(TreeNode root) {

        if(root==null) return true;

        if(depth(root)==-1) return false;

        return true;

        

       

        

        

        

    }

    

    public int depth(TreeNode node)

    {

        if(node==null) return 0;

        int d1=depth(node.right);

        int d2=depth(node.left);

        if(d1==-1||d2==-1) return -1;

        if(Math.abs(d1-d2)>1) return -1;

        

    

        

        

        return  Math.max(d1,d2)+1;

        

    }

    

}