一、题目描述
给定两个二叉树，编写一个函数来检验它们是否相同。
如果两个树在结构上相同，并且节点具有相同的值，则认为它们是相同的。
二、题解
2.1深度递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if(p == null && q == null){
return true;
}
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);
}
}
}
2.2拓展
二叉树系列问题模板：
- 相同的树
- 删除二叉搜索树中的节点
- 二叉搜索树中的插入操作
- 二叉搜索树中的搜索
- 验证二叉搜索树
参考：LeetCode题解
2.2.1二叉树
- 二叉树的节点值加一、判断两颗二叉树是否相等
2.2.2BST二叉搜索树
删除操作设计情况较复杂，不在举栗
2.3广度队列、前、中、后单栈、后序双栈解法
/**
* 递归解法
*
* @param p
* @param q
* @return
*/
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null) return true;
if (p == null) return false;
if (q == null) return false;
if (p.val == q.val) {
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
} else {
return false;
}
}
/**
* 非递归前序遍历判断相等
* 根左右
*
* @param p
* @param q
* @return
*/
public boolean isSameTreePre(TreeNode p, TreeNode q) {
if (p == null && q == null) return true;
if (p == null) return false;
if (q == null) return false;
Stack<TreeNode> stackP = new Stack<>();
stackP.add(p);
Stack<TreeNode> stackQ = new Stack<>();
stackQ.add(q);
while (!stackP.isEmpty() || !stackQ.isEmpty()) {
TreeNode pNode = stackP.pop();
TreeNode qNode = stackQ.pop();
if (pNode == null && qNode == null) continue;
if (pNode == null) return false;
if (qNode == null) return false;
if (pNode.val != qNode.val) {
return false;
} else {
stackP.add(pNode.right);
stackQ.add(qNode.right);
stackP.add(pNode.left);
stackQ.add(qNode.left);
}
}
return true;
}
/**
* 非递归中序遍历判断两棵二叉树是否相等
* 左根右
*
* @param p
* @param q
* @return
*/
public boolean isSameTreeMid(TreeNode p, TreeNode q) {
if (p == null && q == null) return true;
if (p == null) return false;
if (q == null) return false;
Stack<TreeNode> stackP = new Stack<>();
Stack<TreeNode> stackQ = new Stack<>();
while (p != null || q != null || !stackP.isEmpty() || !stackQ.isEmpty()) {
if (p != null && q != null) {
stackP.add(p);
stackQ.add(q);
p = p.left;
q = q.left;
} else if (q == null && p == null) {
p = stackP.pop();
q = stackQ.pop();
if (p.val != q.val) {
return false;
}
p = p.right;
q = q.right;
} else {
return false;
}
}
return true;
}
/**
* 二叉树后序遍历判断两颗树相等
* 左右根
* 双栈解法
*
* @param p
* @param q
* @return
*/
public boolean isSameTreeEnd1(TreeNode p, TreeNode q) {
if (p == null && q == null) return true;
if (p == null) return false;
if (q == null) return false;
Stack<TreeNode> stackP = new Stack<>();
Stack<TreeNode> stackPTmp = new Stack<>();
Stack<TreeNode> stackQ = new Stack<>();
Stack<TreeNode> stackQTmp = new Stack<>();
while (p != null || q != null || !stackPTmp.isEmpty() || !stackQTmp.isEmpty()) {
if (p != null && q != null) {
stackP.add(p);
stackPTmp.add(p);
stackQ.add(q);
stackQTmp.add(p);
p = p.right;
q = q.right;
} else if (p == null && q == null) {
p = stackPTmp.pop();
p = p.left;
q = stackQTmp.pop();
q = q.left;
} else {
return false;
}
}
while (!stackP.isEmpty() && !stackQ.isEmpty()) {
if (stackP.pop().val != stackQ.pop().val) {
return false;
}
}
return true;
}
/**
* 二叉树后序遍历单栈实现对比判断
*
* @param p
* @param q
* @return
*/
public boolean isSameTreeEnd2(TreeNode p, TreeNode q) {
if (p == null && q == null) return true;
if (p == null) return false;
if (q == null) return false;
Stack<TreeNode> stackP = new Stack<>();
stackP.add(p);
TreeNode pPre = null;
Stack<TreeNode> stackQ = new Stack<>();
stackQ.add(q);
TreeNode qPre = null;
while (!stackP.isEmpty() || !stackQ.isEmpty()) {
p = stackP.peek();
q = stackQ.peek();
if (((pPre != null && (p.left == pPre || p.right == pPre)) || (p.left == null && p.right == null)) &&
((qPre != null && (q.left == qPre || q.right == qPre)) || (q.left == null && q.right == null))) {
if (p.val != q.val) {
return false;
}
pPre = p;
qPre = q;
stackP.pop();
stackQ.pop();
} else {
if (p.right != null && q.right != null) {
stackP.add(p.right);
stackQ.add(q.right);
} else if (p.right != null || q.right != null) {
return false;
}
if (p.left != null && q.left != null) {
stackP.add(p.left);
stackQ.add(q.left);
} else if (p.left != null || q.left != null) {
return false;
}
}
}
return true;
}