101. Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following [1,2,2,null,3,null,3] is not: 1 / \ 2 2 \ \ 3 3 Note: Bonus points if you could solve it both recursively and iteratively.
L: mirror tree:
Node mirror(Node node) { if (node == null) return node; /* do the subtrees */ Node left = mirror(node.left); Node right = mirror(node.right); /* swap the left and right pointers */ node.left = right; node.right = left; return node; }
Traverse both left and right branches of the root symmetricaly and check if the values are equal.
想testcase, 如何遍历, 与谁比较, 是否为空,
就是判断值和关系是否相等
先序遍历
public boolean isSymmetric(TreeNode root) { if (root == null) { return true; } Stack<TreeNode> left = new Stack<>(); Stack<TreeNode> right = new Stack<>(); if (root.left != null && root.right == null || root.right != null && root.left == null) { return false; } if (root.left == null && root.right == null) { return true; } left.push(root.left); right.push(root.right); while (!left.isEmpty() && !right.isEmpty()) { TreeNode cur1 = left.pop(); TreeNode cur2 = right.pop(); if (cur1.val != cur2.val) { return false; } if (cur1.left != null) { left.push(cur1.left); } if (cur2.right != null) { right.push(cur2.right); } if (left.size() != right.size()) return false; if (cur2.left != null) { left.push(cur2.left); } if (cur1.right != null) { right.push(cur1.right); } if (left.size() != right.size()) return false; } return left.size() == right.size(); }
递归
public boolean isSymmetric(TreeNode root) { if(root == null){ return true; } return helper(root.left, root.right); } public boolean helper(TreeNode p, TreeNode q){ if(p == null && q == null) { return true; } if(p == null || q == null) { return false; } return ( p.val == q.val && helper(p.left, q.right) && helper(p.right, q.left) ); }