[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.

https://oj.leetcode.com/problems/balanced-binary-tree/

思路：注意不要写成top-down的顺序，会造成多次遍历树，应该采用bottom up遍历的顺序依次判断。

　　注意java的没有引用传递，所以用了一个wrap类。

复杂度：top-down: 时间NlogN，空间N；  bottom-up：时空N。

/*
 *  be careful about the order of traversal!!!don't do recursion in recursion that resulting traverse the tree many times.
 *      so we need to return the height and isBalanced of a node at the same time, but we can only return one, so if we use C++
 *      we can use reference, but it's java... so we have to wrap a primitive or using an array.
 * https://oj.leetcode.com/discuss/3931/can-we-have-a-better-solution
 * http://www.cnblogs.com/remlostime/archive/2012/10/27/2742987.html
 * 
 */
public class Solution {

    public boolean isBalanced(TreeNode root) {
        Height height = new Height();
        return checkBalanced(root, height);
    }

    private boolean checkBalanced(TreeNode root, Height h) {
        if (root == null) {
            h.height = 0;
            return true;
        }
        Height lh = new Height();
        Height rh = new Height();
        boolean leftBalanced = checkBalanced(root.left, lh);
        boolean rightBalanced = checkBalanced(root.right, rh);
        h.height = Math.max(lh.height, rh.height) + 1;

        return leftBalanced && rightBalanced && (Math.abs(lh.height - rh.height) <= 1);
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(5);
        root.right = new TreeNode(7);
        root.right.left = new TreeNode(3);
        root.right.left.right = new TreeNode(3);
        root.left = new TreeNode(3);

        System.out.println(new Solution().isBalanced(root));
    }

    private static class Height {
        int height;
    }

}

第二遍记录: 此题需要同时返回两个参数，加上java无引用传递，只能用类包装下。

public class Solution {
    public boolean isBalanced(TreeNode root) {
        Height height = new Height();
        return check(root,height);
    }
    
    private boolean check(TreeNode root,Height h){
        if(root==null){
            h.height=0;
            return true;
        }
        Height lh = new Height();
        Height rh = new Height();
        
        boolean leftBalanced= check(root.left,lh);
        boolean rightBalanced = check(root.right,rh);
        h.height=Math.max(lh.height,rh.height)+1;
        
        return leftBalanced && rightBalanced && (Math.abs(lh.height-rh.height)<2);
    }
    
    private static class Height{
        int height;
    }
}

第三遍记录：用数据也可以模拟引用传递。

　　此题需要采用后续遍历，并同时返回高度和是否平衡，所以需要模拟引用传递的形式。

public class Solution {
    
    public boolean isBalanced(TreeNode root) {
        return check(root,new int[1]);        
    }
    
    private boolean check(TreeNode root,int[] height){
        if(root==null){
            height[0]=0;
            return true;        
        }
        int[] lh = new int[1];
        int[] rh = new int[1];
        boolean lb= check(root.left,lh);
        boolean rb= check(root.right,rh);
        if(!lb||!rb)
            return false;
        height[0]= Math.max(lh[0],rh[0])+1;
        
        return !(Math.abs(lh[0]-rh[0])>1);
        
    }
}