标签：row 先序遍历 mos tle and tput view ast for
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.
现在开始考bfs 了吗, 根据size办事,分情况了
public int findBottomLeftValue(TreeNode root) {
if (root == null) return -1;
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int ans = 1;
while (!q.isEmpty()) {
int size = q.size();
for (int i = 0; i < size; i++) {
TreeNode cur = q.poll();
if (cur.left != null) {
q.offer(cur.left);
}
if (cur.right != null) {
q.offer(cur.right);
}
if (i == 0) {
ans = cur.val;
}
}
}
return ans;
}
dfs: 先序遍历 + 树的深度, 跟此题类似: 199 Binary Tree Right Side View
public class Solution {
int deep = 0;
int ans = -1;
public int findBottomLeftValue(TreeNode root) {
if (root == null) return -1;
//int curDeep = 0;
dfs(root, 0);
return ans;
}
private void dfs(TreeNode root, int curDeep) {
if (root == null) {
return;
}
if (root.left == null && root.right == null && deep == curDeep) {
deep++;
ans = root.val;
return;
}
if (deep == curDeep) {
deep++;
//curDeep++;
}
dfs(root.left, curDeep + 1);
dfs(root.right, curDeep + 1);
}
}
513. Find Bottom Left Tree Value
标签：row 先序遍历 mos tle and tput view ast for
原文地址：http://www.cnblogs.com/apanda009/p/7295517.html
