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LeetCode-100-Same-Tree.java
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LeetCode-100-Same-Tree.java
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/*
LeetCode: https://leetcode.com/problems/same-tree/
LintCode:
JiuZhang:
ProgramCreek: http://www.programcreek.com/2012/12/check-if-two-trees-are-same-or-not/
Analysis:
1.DFS: Tooooo easy.
2.BFS
Using recursive approach, we need two stach for the two Tree. The order doesn't matter, just need to Traversal all the nodes.
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
// 1.DFS
// 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 false;
// return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
// }
// 2.BFS (similar to inorder Traversal)
// Similar to: https://leetcode.com/problems/same-tree/discuss/281890/Clean-Java-Non-recursive-Solution
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 false;
Stack<TreeNode> first = new Stack<>();
Stack<TreeNode> second = new Stack<>();
first.push(p);
second.push(q);
while(!first.isEmpty() && !second.isEmpty()) {
p = first.pop();
q = second.pop();
if (p == null && q == null) {
// do nothing
} else if (p == null || q == null) {
return false;
} else {
if (p.val != q.val) {
return false;
}
// pushing left node, or right node, the order doesn't matter.
first.push(p.left);
second.push(q.left);
first.push(p.right);
second.push(q.right);
}
}
return true;
}
}