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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