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.
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var findBottomLeftValue = function(root) {
var res=new Array();
var queue=new Array();
var bottomLeft;
queue.push(root);
while(queue.length>0){
var temp=new Array();
while(queue.length>0){
var n=queue.shift();
bottomLeft=n.val;
if(n.right!=null){
temp.push(n.right);
}if(n.left!=null){
temp.push(n.left);
}
}
queue=temp;
}
return bottomLeft;
};
一道BFS题目,思路就是用一个队列,遍历每层。