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Find Bottom Left Tree Value |找到二叉树最后一层的最左边的节点的值

2017-03-11 15:05 513 查看
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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思路:层次遍历二叉树,并在遍历时记录下一层的个数和当前的层的个数,当当前层的数量为0是,如果下一层的节点数量大于0,则更新value的值为队列中队首的值。

/**
* 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 findBottomLeftValue(TreeNode root) {
int value = root.val;
int last = 1, next = 0;
LinkedList<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {

TreeNode node = queue.poll();
if (node.left != null) {
queue.addLast(node.left);
next++;
}
if (node.right != null) {
queue.add(node.right);
next++;
}
last--;
if (last == 0) {
if (next != 0) {
value = queue.peek().val;
}
last = next;
next = 0;
}

}
return value;
}
}




看了答案,广度遍历时,先插入右孩子,然后再插入左孩子,这样最后的一个节点就是要找的节点。
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