Leetcode 110. Balanced Binary Tree

方法1: top-down recursion。这是最简单直白的。对于当前节点，我们检查他的左右子树是不是平衡二叉树，并且检查他左右子树高度差是不是小于等于1。时间复杂avergae case nlogn worst case n^2，空间复杂n。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        if(root == null) return true;
        if(root.left == null && root.right == null) return true;
        TreeNode left = root.left;
        TreeNode right = root.right;
        int heightLeft = heightOfTree(left);
        int heightRight = heightOfTree(right);
        boolean booleanLeft = isBalanced(left);
        boolean booleanRight = isBalanced(right);
        return booleanLeft&&booleanRight&&(Math.abs(heightLeft-heightRight)<=1);
    }
    
    private int heightOfTree(TreeNode root){
        if(root == null) {
            return 0;
        }
        TreeNode left = root.left;
        TreeNode right = root.right;
        return 1 + Math.max(heightOfTree(left),heightOfTree(right));
    }
}

方法2: bottom-up recursion。时间复杂n，空间复杂n。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        return getHeight(root) != -1;
    }
    
    private int getHeight(TreeNode node) {
        if (node == null) return 0;

        int left = getHeight(node.left);
        int right = getHeight(node.right);

        // left, right subtree is unbalanced or cur tree is unbalanced
        if (left == -1 || right == -1 || Math.abs(left - right) > 1) return -1;

        return Math.max(left, right) + 1;
    }
}

总结：

• 这道题目的复杂度分析真的好难，我尝试着不用数学去分析，但是没有搞明白。