[leetcode]解决Same Tree的一点小心得

本次选择的题目是

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.

Solution：

首先我想到一个直接的方法，利用栈，按同样的顺序将p和q的各个节点push进去，再pop出来对比它们的val值，当然不是先全部push完再pop，而是先左再右，pop出来左节点，然后左节点一样了，push左节点的两个子节点，再pop出来右节点，检测右节点的val，push右节点的两个子节点，然后栈里还有左节点的两个子节点和右节点的两个子节点（假设它有，这里有4种情况，都没有，都有，都有左子节点，都有右子节点）一一对比。

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if(p == null && q == null){
return true;
}else if(p != null && q != null){
Stack<TreeNode> stack1 = new Stack<TreeNode>();
Stack<TreeNode> stack2 = new Stack<TreeNode>();
stack1.push(p);
stack2.push(q);
while(!stack1.empty() && !stack2.empty()){
TreeNode temp1 = stack1.pop();
TreeNode temp2 = stack2.pop();
if(temp1.val!=temp2.val){
return false;
}
if(temp1.right!=null&&temp1.left!=null&&temp2.right!=null&&temp2.left!=null){
stack1.push(temp1.left);
stack1.push(temp1.right);
stack2.push(temp2.left);
stack2.push(temp2.right);
}else if(temp1.right!=null&&temp1.left==null&&temp2.right!=null&&temp2.left==null){
stack1.push(temp1.right);
stack2.push(temp2.right);
}else if(temp1.right==null&&temp1.left!=null&&temp2.right==null&&temp2.left!=null){
stack1.push(temp1.left);
stack2.push(temp2.left);
}else if(temp1.right==null&&temp1.left==null&&temp2.right==null&&temp2.left==null){

}else{
return false;
}
}
return true;
}else{
return false;
}
}
}

discussion中提到的简单方法：

public boolean isSameTree(TreeNode p, TreeNode q) {
if(p == null && q == null) return true;
if(p == null || q == null) return false;
if(p.val == q.val)
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
return false;
}