力扣labuladong——一刷day45

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前言

二叉树的递归分为「遍历」和「分解问题」两种思维模式，这道题需要用到「遍历」的思维

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一、力扣270. 最接近的二叉搜索树值

/**
 * 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 {
    double flag = Integer.MAX_VALUE;
    double res = 0;
    public int closestValue(TreeNode root, double target) {
        fun(root,target);
        return (int)res;
    }
    public void fun(TreeNode root , double target){
        if(root == null){
            return;
        }
        fun(root.left,target);
        double temp = Math.abs(target - root.val);
        if(temp < flag){
            flag = temp;
            res = root.val;
        }
        fun(root.right,target);
    }
}
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二、力扣404. 左叶子之和

/**
 * 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 {
    int sum = 0;
    public int sumOfLeftLeaves(TreeNode root) {
        fun(root);
        return sum;
    }
    public void fun(TreeNode root){
        if(root == null){
            return;
        }
        if(root.left != null){
            if(root.left.left == null && root.left.right == null){
                sum += root.left.val;
            }
        }
        fun(root.left);
        fun(root.right);
    }
}
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三、力扣617. 合并二叉树

/**
 * 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 TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        return fun(root1,root2);
    }
    public TreeNode fun(TreeNode root1, TreeNode root2){
        if(root1 != null && root2 != null){
            root1.val = root1.val + root2.val;
        }else if(root1 == null && root2 != null){
            return root2;
        }else if(root1 != null && root2 == null){
            return root1;
        }else{
            return null;
        }
        TreeNode l = fun(root1.left, root2.left);
        TreeNode r = fun(root1.right, root2.right);
        root1.left = l;
        root1.right = r;
        return root1;
    }
}
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四、力扣623. 在二叉树中增加一行

/**
 * 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 {
    int len = 0;
    public TreeNode addOneRow(TreeNode root, int val, int depth) {
        if(depth == 1){
            return new TreeNode(val,root,null);
        }
        len = depth-1;
        return fun(root, val, 1);
    }
    public TreeNode fun(TreeNode root, int val, int depth){
        if(root == null){
            return null;
        }
        if(depth == len){
            root.left = new TreeNode(val,root.left,null);
            root.right = new TreeNode(val,null,root.right);
            return root;
        }
        TreeNode l = fun(root.left, val, depth+1);
        TreeNode r = fun(root.right, val, depth +1);
        root.left = l;
        root.right = r;
        return root;
    }
}
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