[LeetCode]513. Find Bottom Left Tree Value

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本文介绍两种方法来解决寻找二叉树最底层最左侧节点值的问题。方法一采用标准层序遍历实现，而方法二通过调整节点入队顺序确保最终队列中的节点即为所求。

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:

Note: You may assume the tree (i.e., the given root node) is not NULL.

方法一：中规中矩的层序遍历

/**
 * 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 res;
        Queue<TreeNode> que = new LinkedList<TreeNode>();
        que.offer(root);
        TreeNode leftmost=null;
        TreeNode temp=null;
        int levelc=1; //marks the number of node in one level
        int levelm=0;
        
        while(!que.isEmpty()){
            leftmost=que.element();
            while(levelc>0){
                temp = que.poll();
                levelc--;
                if(temp.left!=null) {que.offer(temp.left);levelm++;}
                if(temp.right!=null) {que.offer(temp.right);levelm++;}
            }
            levelc=levelm;
            levelm=0;
        }
        
        
        res=leftmost.val;
        
        return res;
    }
}

方法二：左右反一下，进入队列的时候，先压右子树，再压左子树，那么最后留在queue里面的一定是最左边的节点

public class Solution {  
    public int findBottomLeftValue(TreeNode root) {  
        Queue<TreeNode> q = new LinkedList<>();  
        q.add(root);  
        while (!q.isEmpty()) {  
            root = q.poll();  
            if (root.right != null)  
                q.add(root.right);  
            if (root.left != null)  
                q.add(root.left);  
        }  
        return root.val;  
    }  
  
}