平衡二叉树

平衡二叉树的定义:

如果每一个节点的左子树和右子树的高度差不超过1,那么这可树就是平衡二叉树。

判断一个树是否为平衡树,把握三点。

(1)左子树是否平衡 (2)右子树是否平衡 (3)左子树的高度和右子树的高度差值低于1

递归的套路

package class_04;

public class Code_06_IsBalancedTree {

	public static class Node {
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
			this.value = data;
		}
	}

	public static class ReturnData{
		public boolean isB;   // 是否平衡
		public int h;         // 该节点的高度
		public ReturnData(boolean isB, int h) {
			this.isB = isB;
			this.h = h;
		}
	}
	
	/*
	 * 思路:判断一个树是否为平衡树,(1)左子树是否平衡 (2)右子树是否平衡 (3)左子树的高度和右子树的高度差值低于1
	 * */
	public static ReturnData process(Node node) {
		if(node == null) {                       // 判断是否为空,空树是平衡的
			return new ReturnData(true, 0);
		}
		
		ReturnData leftData= process(node.left); // 判断左子树是否平衡
		if(!leftData.isB) {
			return new ReturnData(false, 0);
		}
		
		ReturnData rightData= process(node.left); // 判断右子树是否平衡
		if(!rightData.isB) {
			return new ReturnData(false, 0);
		}
		
		if(Math.abs(leftData.h - rightData.h) > 1) { // 如果左子树和右子树的高度值差超过1
			return new ReturnData(false, 0);
		}
		
		return new ReturnData(true, Math.max(leftData.h, rightData.h) + 1);  // 往下走高度+1
	}
	
	public static boolean isBalance(Node head) {
		boolean[] res = new boolean[1];
		res[0] = true;
		getHeight(head, 1, res);
		return res[0];
	}

	public static int getHeight(Node head, int level, boolean[] res) {
		if (head == null) {
			return level;
		}
		int lH = getHeight(head.left, level + 1, res);
		if (!res[0]) {
			return level;
		}
		int rH = getHeight(head.right, level + 1, res);
		if (!res[0]) {
			return level;
		}
		if (Math.abs(lH - rH) > 1) {
			res[0] = false;
		}
		return Math.max(lH, rH);
	}

	public static void main(String[] args) {
		Node head = new Node(1);
		head.left = new Node(2);
		head.right = new Node(3);
		head.left.left = new Node(4);
		head.left.right = new Node(5);
		head.right.left = new Node(6);
		head.right.right = new Node(7);

		System.out.println(isBalance(head));

	}

}

 

posted @ 2019-01-15 23:29  Horken  阅读(135)  评论(0编辑  收藏  举报