LeetCode 662. 二叉树最大宽度
题目描述
思路分析
层序遍历
参考代码
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func widthOfBinaryTree(root *TreeNode) int {
if root == nil {
return 0
}
res := 0
root.Val = 1
queue := []*TreeNode{root}
for len(queue) > 0 {
size := len(queue)
first, last := 0, 0
for i := 0; i < size; i++ {
node := queue[0]
if i == 0 {
first = node.Val
}
if i == size-1 {
last = node.Val
}
queue = queue[1:]
if node.Left != nil {
node.Left.Val = node.Val*2 - 1
queue = append(queue, node.Left)
}
if node.Right != nil {
node.Right.Val = node.Val * 2
queue = append(queue, node.Right)
}
}
res = max(res, last-first+1)
}
return res
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
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class Solution {
public int widthOfBinaryTree(TreeNode root) {
int res = 0;
if (root == null) {
return res;
}
Queue<TreeNode> queue = new LinkedList<>();
root.val = 1;
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
int start = 0, end = 0;
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
if (i == 0) {
start = node.val;
}
if (i == size - 1) {
end = node.val;
}
if (node.left != null) {
node.left.val = node.val * 2 - 1;
queue.offer(node.left);
}
if (node.right != null) {
node.right.val = node.val * 2;
queue.offer(node.right);
}
}
res = Math.max(res, end - start + 1);
}
return res;
}
}
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