/*
* 面试题55-题目二:平衡二叉树
* 题目:输入一棵二叉树,判断该二叉树是否是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1,那么它就是一棵平衡二叉树
* 思路:每个节点只遍历一次,采用后序遍历的方式遍历二叉树的每个节点
*/
public class No55IsBalanced_Solution {
public static void main(String[] args) {
No55IsBalanced_Solution n = new No55IsBalanced_Solution();
TreeNode10 root = new TreeNode10(1);
TreeNode10 p2 = new TreeNode10(2);
TreeNode10 p3 = new TreeNode10(3);
TreeNode10 p4 = new TreeNode10(4);
TreeNode10 p5 = new TreeNode10(5);
TreeNode10 p6 = new TreeNode10(6);
TreeNode10 p7 = new TreeNode10(7);
TreeNode10 p8 = new TreeNode10(8);
TreeNode10 p9 = new TreeNode10(9);
root.left = p2;
root.right = p3;
p2.left = p4;
p2.right = p5;
p4.left = p7;
p4.right = p8;
p7.left = p9;
p3.right = p6;
System.out.print(n.IsBalanced_Solution(root));
}
public boolean IsBalanced_Solution(TreeNode10 root) {
if (root == null ) {
return true;
}
int left = getDepth(root.left);
int right = getDepth(root.right);
int diff = left - right;
if (diff >= -1 && diff <= 1) {
return true;
} else {
return false;
}
}
private int getDepth(TreeNode10 root) {
if (root == null) {
return 0;
}
int depth = 0;
int leftNode = getDepth(root.left);
int rightNode = getDepth(root.right);
depth = leftNode > rightNode ? leftNode : rightNode;
return depth + 1;
}
}
面试题55-题目二:平衡二叉树
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转载自blog.csdn.net/juaner1993/article/details/82760680
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