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Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
题意:判断两棵树是否一样。思路:一种是递归,一种是利用队列做层次遍历判断。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null) return true;
else if (p != null && q == null) return false;
else if (p == null && q != null) return false;
else {
if (p.val != q.val)
return false;
else return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}
}/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private boolean isSameNode(TreeNode p, TreeNode q) {
if (p == null && q == null) return true;
if ((p != null && q == null) || (p == null && q != null) || (p.val != q.val))
return false;
return true;
}
public boolean isSameTree(TreeNode p, TreeNode q) {
if (!isSameNode(p, q)) return false;
if (p == null && q == null) return true;
LinkedList<TreeNode> left = new LinkedList<TreeNode>();
LinkedList<TreeNode> right = new LinkedList<TreeNode>();
left.add(p);
right.add(q);
while (left.size() != 0 && right.size() != 0) {
TreeNode ltmp = left.peekFirst();
TreeNode rtmp = right.peekFirst();
left.removeFirst();
right.removeFirst();
if (!isSameNode(ltmp.left, rtmp.left)) return false;
if (ltmp.left != null && rtmp.left != null) {
left.add(ltmp.left);
right.add(rtmp.left);
}
if (!isSameNode(ltmp.right, rtmp.right)) return false;
if (ltmp.right != null && rtmp.right != null) {
left.add(ltmp.right);
right.add(rtmp.right);
}
}
return true;
}
}
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原文地址:http://blog.csdn.net/u011345136/article/details/45287589