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| class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
TreeNode tmp = root.left;
root.left = root.right;
root.right = tmp;
invertTree(root.left);
invertTree(root.right);
return root;
}
}
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DFS
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| class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
invertTree(root.left);
invertTree(root.right);
TreeNode tmp = root.left;
root.left = root.right;
root.right = tmp;
return root;
}
}
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DFS
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| class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
TreeNode tmp = root.left;
root.left = invertTree(root.right);
root.right = invertTree(tmp);
return root;
}
}
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BFS
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| class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
if (node.left != null) {
queue.add(node.left);
}
if (node.right != null) {
queue.add(node.right);
}
TreeNode tmp = node.left;
node.left = node.right;
node.right = tmp;
}
return root;
}
}
|
References
226. Invert Binary Tree
剑指 Offer 27. 二叉树的镜像