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class Solution {
private List<Integer> leftBound;
private List<Integer> rightBound;
private List<Integer> leaves;
public List<Integer> boundaryOfBinaryTree(TreeNode root) {
if (root == null)
return new ArrayList<>();
leftBound = new ArrayList<>();
rightBound = new ArrayList<>();
leaves = new ArrayList<>();
List<Integer> ans = new ArrayList<>();
ans.add(root.val);
if (root.left != null) {
dfs(root.left, -1);
}
if (root.right != null) {
dfs(root.right, 1);
}
for (int node : leftBound) {
ans.add(node);
}
for (int node : leaves) {
ans.add(node);
}
for (int node : rightBound) {
ans.add(node);
}
return ans;
}
private void dfs(TreeNode node, int status) {
if (node == null)
return;
if (node.left == null && node.right == null) {
leaves.add(node.val);
} else if (status == -1) {
leftBound.add(node.val);
if (node.left != null) {
dfs(node.left, -1);
dfs(node.right, 0);
} else {
dfs(node.right, -1);
}
} else if (status == 1) {
rightBound.add(0, node.val);
if (node.right != null) {
dfs(node.left, 0);
dfs(node.right, 1);
} else {
dfs(node.left, 1);
}
} else {
dfs(node.left, 0);
dfs(node.right, 0);
}
}
}
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我们dfs然后告诉每个node它是左边界, 右边界还是都不是. 通过判断是否有children来判断是否是leaf.
时间复杂度: O(n)
空间复杂度: O(n)
