LeetCode: isSameTree1 解题报告

isSameTree1

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.

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
// solution 1:
public boolean isSameTree1(TreeNode p, TreeNode q) {
if (p == null && q == null) {
return true;
}

if (p == null || q == null) {
return false;
}

return p.val == q.val &&
isSameTree(p.left, q.left) &&
isSameTree(p.right, q.right);
}

// Solution 2:
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.push(p);
s2.push(q);

while (!s1.isEmpty() && !s2.isEmpty()) {
TreeNode cur1 = s1.pop();
TreeNode cur2 = s2.pop();

// 弹出的节点的值必须相等
if (cur1.val != cur2.val) {
return false;
}

// tree1的right节点，tree2的right节点，可以同时不为空，也可以同时为空，否则返回false.
if (cur1.left != null && cur2.left != null) {
s1.push(cur1.left);
s2.push(cur2.left);
} else if (!(cur1.left == null && cur2.left == null)) {
return false;
}

// tree1的左节点，tree2的left节点，可以同时不为空，也可以同时为空，否则返回false.
if (cur1.right != null && cur2.right != null) {
s1.push(cur1.right);
s2.push(cur2.right);
} else if (!(cur1.right == null && cur2.right == null)) {
return false;
}
}

return true;
}
}

View Code

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/IsSameTree1.java