[刷题]Balanced Binary Tree

原创于 2015-03-23 17:03:30 发布 · 360 阅读

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本文介绍了一种高效判断二叉树是否平衡的方法。通过递归辅助函数helper，该算法能够计算每个节点的左右子树高度差，并以此来确定整棵树是否平衡。如果任意两棵子树的高度差超过1，则认为这棵树不平衡。

[LintCode]Balanced Binary Tree

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    public boolean isBalanced(TreeNode root) {
        // 2015-3-23
        return helper(root) != -1;
    }

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

    	// divide
    	int left = helper(root.left);
    	int right = helper(root.right);

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