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
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Given a binary tree, find the leftmost value in the last row of the tree.
Example 1:
Example 2:
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。
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。
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.
Subscribe to see which companies asked this question.
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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