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513. Find Left Most Element- User Accepted: 870
- User Tried: 943
- Total Accepted: 882
- Total Submissions: 2399
- Difficulty: Medium
Given a binary tree, find the leftmost value in the last row of the tree.
Example 1:
Input:
2
/ \
1 3
Output:
1
Example 2:
Input:
1
/ \
2 3
/ / \
4 5 6
/
7
Output:
7
Note: You may assume the tree (i.e., the given root node) is not NULL.
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BFS的变形题，掌握了BFS这题难度其实是EASY
变化就是记录一下每一层的第一个元素，循环的最后自然是最后一层的最左边元素
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int findLeftMostNode(TreeNode root) {
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
TreeNode res = null;
while(!queue.isEmpty()){
int breadth = queue.size();
for(int i = 0; i < breadth; i++){
TreeNode cur = queue.poll();
if(i == 0)
res = cur;
if(cur.left != null)
queue.offer(cur.left);
if(cur.right != null)
queue.offer(cur.right);
}
}
return res.val;
}
}再来一个recursive的DFS写法
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
int maxDepth = 0;
int res = 0;
public int findLeftMostNode(TreeNode root) {
helper(root, 1);
return res;
}
private void helper(TreeNode cur, int depth){
if(depth > maxDepth){//第一次进入下一层必然是当层的最左节点，记录这个节点的val
maxDepth = depth;
res = cur.val;
}
if(cur.left != null) //注意先遍历左节点
helper(cur.left, depth + 1);
if(cur.right != null)
helper(cur.right, depth + 1);
}
}