【LeetCode】Same Tree

题目

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.

解答

判断两棵树是否相同。

递归，先判断根节点是否相同，如果相同，继续递归判断左右节点；

迭代，利用栈，当两棵树都不为空，并且节点值相同时，分别对左右子节点入栈，循环判断。代码如下：

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } *///递归解法，如果根节点相同，再递归的看左子树和右子树是否相同public class Solution {    public boolean isSameTree(TreeNode p, TreeNode q) {        if(p==null&&q==null){            return true;        }else if(p!=null&&q!=null){            if(p.val==q.val)                return isSameTree(p.left,q.left)&&isSameTree(p.right,q.right);            else                return false;        }else{            return false;        }    }}//递归2public boolean isSameTree(TreeNode p, TreeNode q) {    if(p==null&&q==null){        return true;    }    if(p==null&&q!=null||(p!=null&&q==null)){  //  if(p==null||q==null)        return false;    }    if(p.val!=q.val){        return false;    }    return isSameTree(p.left,q.left)&&isSameTree(p.right,q.right);}// 迭代，使用栈public boolean isSameTree(TreeNode p,TreeNode q){    if(p==null&&q==null){        return true;    }    if(p==null||q==null){        return false;    }    Stack<TreeNode> s1=new Stack<TreeNode>();    Stack<TreeNode> s2=new Stack<TreeNode>();    s1.add(p);    s2.add(q);    while(!s1.isEmpty()&&!s2.isEmpty()){        TreeNode t1=s1.pop();        TreeNode t2=s2.pop();        if(t1==null&&t2==null){            continue;        }else if(t1!=null&&t2!=null&&t1.val==t2.val){  //注意入栈顺序            s1.push(t1.right);            s1.push(t1.left);            s2.push(t2.right);            s2.push(t2.left);        }else{            return false;        }    }    return true;}

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