leetcode110. Balanced Binary Tree

110. 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.

解法

分治，判断左子树是不是平衡树，判断右子树是不是平衡树。

/**
 * Definition for a binary tree node.
 * 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;
        }

        boolean left = isBalanced(root.left);
        boolean right = isBalanced(root.right);
        if (Math.abs(maxDepth(root.left) - maxDepth(root.right)) > 1) {
            return false;
        }
        if (!left || !right) {
            return false;
        }

        return true;
    }
    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }

        int left = maxDepth(root.left) + 1;
        int right = maxDepth(root.right) + 1;

        return left > right ? left : right;
    }
}

解法二

解法一效率太低，是由于进行了两层分治算法，其实只用一层就可以，在计算每一层的深度时，判断是否相差大于1。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isBalanced(TreeNode root) {
        return maxDepth(root) != -1;
    }

    private int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }

        int left = maxDepth(root.left);
        int right = maxDepth(root.right);
        if (left == -1 || right == -1 || Math.abs(left-right) > 1) {
            return -1;
        }
        return Math.max(left, right) + 1;
    }
}