4.＜tag-二叉树和基本性质＞lt.404. 左叶子之和+ lt.513. 找树左下角的值 dbc

lt.404. 左叶子之和

[案例需求]

[思路分析一, 递归法]

[代码实现]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    // 递归函数: 参数是被看做是根节点的当前节点, 返回值是当前加点的左孩子结点的和
    public int sumOfLeftLeaves(TreeNode root) {
        //递归出口(当前节点为叶结点了, 只能退出了)
        if(root == null)return 0;

        //单层逻辑
        int sum = 0;

        if(root.left != null && root.left.left == null && root.left.right == null)
            sum += root.left.val;

        sum += sumOfLeftLeaves(root.left);
        sum += sumOfLeftLeaves(root.right);
        //返回值;
        return sum;
    }
}

[思路分析二, 迭代法]

• 使用层序遍历遍历每一层, 当从队列中取出一个node时(我们看做是root), 如果这个node存在左叶子结点, 直接把他的值叠加到sum中, 然后把node的左右子树放入队列, 以进行下一层的迭代;

[代码实现]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {

    public int sumOfLeftLeaves(TreeNode root) {
       //层序迭代法
       Queue<TreeNode> queue = new LinkedList<>();
       if(root == null)return 0;
       queue.add(root);

       int sum = 0;
       while(!queue.isEmpty()){
           int size = queue.size();

           for(int i = 0; i < size; i++){
               root = queue.poll();

               if(root.left != null){
                   if(root.left.left == null && root.left.right == null){
                       sum += root.left.val;
                   }
                   queue.add(root.left);
               }

               if(root.right != null){
                   queue.add(root.right);
               }
           }
       }

       return sum;
    }
}

lt.513. 找树左下角的值

[案例需求]

[思路分析一, 递归]

[代码实现]

• 待补充

[代码实现]

public int findBottomLeftValue(TreeNode root) {
        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);
        TreeNode poll = null;
        while (!deque.isEmpty()){
            poll = deque.poll();
            if (poll.right != null){
                deque.offer(poll.right);
            }
            if (poll.left != null){
                deque.offer(poll.left);
            }
        }
        return poll.val;
    }
}

[思路分析二, 层序遍历修修补补]

[代码实现]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    //函数, 找到最下面的左叶子结点, 返回的是这个节点的值
    public int findBottomLeftValue(TreeNode root) {
        //BFS
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);

         //记录每一层最左边的值
        int ans = 0;

        while(!queue.isEmpty()){
            int size = queue.size(); // 每一层的结点个数 

            for(int i = 0; i < size; i++){
                root = queue.poll();

                //最后一层 
                if(i == 0){
                    ans =  root.val;
                }
                if(root.left != null)queue.add(root.left);
                if(root.right != null)queue.add(root.right);      
            }   
        }
        return ans;
    }
}

[思路分析三, 迭代]

• BFS，题目要求输出最底层最左的元素，我们只需要先序广度搜索这棵树，并把元素分别入队出队，最后出队的那个就是那个最底层最左的元素了。

[代码实现]

public int findBottomLeftValue(TreeNode root) {
        Deque<TreeNode> deque = new LinkedList<>();
        deque.offer(root);
        TreeNode poll = null;
        while (!deque.isEmpty()){
            poll = deque.poll();
            if (poll.right != null){
                deque.offer(poll.right);
            }
            if (poll.left != null){
                deque.offer(poll.left);
            }
        }
        return poll.val;
    }
}